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Gravitation

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Summary

Chapter 7: Gravitation

Summary

  • Introduction to gravitation and its historical context.
  • Kepler's laws of planetary motion:
    • Law of Orbits: Planets move in elliptical orbits with the Sun at one focus.
    • Law of Areas: A line joining a planet to the Sun sweeps out equal areas in equal times.
    • Law of Periods: The square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit.
  • Universal law of gravitation: Describes the gravitational attraction between masses.
  • Gravitational constant (G): A key value in gravitational calculations.
  • Acceleration due to gravity on Earth and variations below and above the surface.
  • Concepts of gravitational potential energy and escape speed.
  • Energy considerations for Earth satellites and their orbits.

Learning Objectives

Learning Objectives

  • Understand the concept of gravitation and its historical context.
  • Explain Kepler's laws of planetary motion.
  • Describe the universal law of gravitation and its implications.
  • Calculate gravitational forces and potential energy in various scenarios.
  • Analyze the motion of satellites and their energy requirements.
  • Apply the concepts of gravitational acceleration at different points relative to the Earth.
  • Solve problems related to escape speed and orbital mechanics.

Detailed Notes

Chapter 7: Gravitation

7.1 Introduction

  • Awareness of attraction towards Earth.
  • Historical context: Galileo's experiments with acceleration due to gravity.

7.2 Kepler's Laws

  1. Law of Orbits: All planets move in elliptical orbits with the Sun at one focus.
  2. Law of Areas: The line joining a planet to the Sun sweeps equal areas in equal times.
  3. Law of Periods: The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit.

7.3 Universal Law of Gravitation

  • Newton's law explaining terrestrial gravitation and Kepler's laws.

7.4 The Gravitational Constant

  • Value: G = 6.67 x 10⁻¹¹ N m² kg⁻².

7.5 Acceleration Due to Gravity of the Earth

  • Constant acceleration experienced by objects towards Earth.

7.6 Acceleration Due to Gravity Below and Above the Surface of Earth

  • Variation of gravitational acceleration with depth and height.

7.7 Gravitational Potential Energy

  • Formula: V(r) = -GMm/r.

7.8 Escape Speed

  • Minimum speed needed to break free from Earth's gravitational influence.

7.9 Earth Satellites

  • Dynamics and energy considerations for satellites orbiting Earth.

7.10 Energy of an Orbiting Satellite

  • Energy calculations for satellites in orbit.

Important Formulas

Physical QuantitySymbolDimensionsUnitRemarks
Gravitational ConstantG[M⁻L L ³⁻²]N m²6.67x10⁻¹¹
Gravitational Potential EnergyV(r)[M L²T⁻²]JGMm (scalar)
Gravitational PotentialU(r)[L²T⁻²]J kg⁻¹GM/r (scalar)
Gravitational IntensityE or g[LT⁻²]m s⁻²GM/r² (vector)

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Gravitation

Common Pitfalls

  • Misunderstanding Kepler's Laws: Students often confuse the laws of planetary motion, especially the law of areas and the law of periods. Ensure you understand that the law of areas states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
  • Forgetting Units: When calculating gravitational forces or potential energy, students frequently forget to keep track of units, leading to incorrect answers. Always check that your units are consistent.
  • Neglecting the Effects of Height: In problems involving escape speed or satellite motion, students may neglect the height of the satellite above the Earth's surface, which can significantly affect calculations.
  • Incorrect Application of Formulas: Students sometimes apply the wrong formula for gravitational force or potential energy. Familiarize yourself with the correct formulas and their conditions.

Tips for Success

  • Draw Diagrams: Visual aids can help in understanding the relationships between different forces and objects in gravitational problems. For example, sketching the orbits of planets can clarify their motion.
  • Practice with Examples: Work through example problems, especially those involving escape speed and gravitational potential energy, to solidify your understanding of the concepts.
  • Review Key Formulas: Make a list of essential formulas related to gravitation, such as the universal law of gravitation and Kepler's laws, and ensure you understand how to derive and apply them.
  • Understand the Concepts: Rather than memorizing formulas, focus on understanding the underlying concepts of gravitation, such as why gravitational forces act the way they do and how they relate to motion.

Practice & Assessment

Multiple Choice Questions

A.

The gravitational force is zero.

B.

The gravitational force is maximum at the center.

C.

The gravitational force is uniform throughout.

D.

The gravitational force increases with distance from the center.
Correct Answer: A

Solution:

The gravitational force inside a spherical shell is zero due to the symmetry of the shell.

A.

It is always positive.

B.

It is zero at the surface of the Earth.

C.

It is negative when defined relative to infinity.

D.

It is independent of the reference point chosen.
Correct Answer: C

Solution:

Gravitational potential energy is defined as negative when the reference point is at infinity, because work is done against the gravitational force to bring a mass from infinity to a point in space.

A.

7.9 km/s

B.

11.2 km/s

C.

9.8 km/s

D.

12.5 km/s
Correct Answer: B

Solution:

The escape speed of a projectile on the Earth's surface is 11.2 km/s.

A.

The satellite will move to a higher circular orbit.

B.

The satellite will move to a lower circular orbit.

C.

The satellite will enter an elliptical orbit with a lower perigee.

D.

The satellite will escape Earth's gravity.
Correct Answer: C

Solution:

Reducing the speed of a satellite in a circular orbit causes it to enter an elliptical orbit with the point of speed reduction becoming the apogee, and the perigee will be lower than the original circular orbit.

A.

It acts only between objects with mass.

B.

It is stronger than electromagnetic force.

C.

It can be shielded by certain materials.

D.

It acts only at short distances.
Correct Answer: A

Solution:

Gravitational force acts between any two objects with mass, regardless of distance, and cannot be shielded.

A.

It would be half the size of Earth's orbit.

B.

It would be twice the size of Earth's orbit.

C.

It would be one-fourth the size of Earth's orbit.

D.

It would be one-eighth the size of Earth's orbit.
Correct Answer: C

Solution:

According to Kepler's third law, T2a3T^2 \propto a^3, where TT is the orbital period and aa is the semi-major axis. If the planet's period is half of Earth's, then (12)2=(aaE)3(\frac{1}{2})^2 = (\frac{a}{a_E})^3, which simplifies to a=12aEa = \frac{1}{\sqrt{2}}a_E. Therefore, the orbital size is one-fourth of Earth's orbit.

A.

It is always attractive.

B.

It is always repulsive.

C.

It can be either attractive or repulsive.

D.

It does not depend on the distance between the masses.
Correct Answer: A

Solution:

The gravitational force between two masses is always attractive, as described by Newton's law of universal gravitation.

A.

Linear momentum

B.

Angular momentum

C.

Gravitational force

D.

Orbital radius
Correct Answer: B

Solution:

Angular momentum is conserved for a planet moving in an elliptical orbit due to the central nature of the gravitational force.

A.

The spacecraft will escape the planet's gravitational influence.

B.

The spacecraft will enter a higher stable orbit.

C.

The spacecraft will crash into the planet.

D.

The spacecraft will maintain its current orbit.
Correct Answer: A

Solution:

Doubling the velocity of a spacecraft in orbit will provide it with energy greater than the escape velocity, allowing it to overcome the gravitational pull of the planet and escape into space.

A.

It increases.

B.

It decreases.

C.

It remains the same.

D.

It becomes zero.
Correct Answer: B

Solution:

The acceleration due to gravity decreases with increasing altitude.

A.

1.76 \times 10^3 \text{ N}

B.

3.52 \times 10^3 \text{ N}

C.

5.28 \times 10^3 \text{ N}

D.

7.04 \times 10^3 \text{ N}
Correct Answer: B

Solution:

The gravitational force can be calculated using the formula: F=GMem(Re+h)2F = \frac{G M_e m}{(R_e + h)^2} where G=6.67×1011 N m2 kg2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}, Me=6.0×1024 kgM_e = 6.0 \times 10^{24} \text{ kg}, m=200 kgm = 200 \text{ kg}, Re=6.4×106 mR_e = 6.4 \times 10^6 \text{ m}, and h=400×103 mh = 400 \times 10^3 \text{ m}. Substituting the values, F=6.67×1011×6.0×1024×200(6.4×106+400×103)2=3.52×103 NF = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{(6.4 \times 10^6 + 400 \times 10^3)^2} = 3.52 \times 10^3 \text{ N}.

A.

Planets move faster when they are closer to the Sun.

B.

Planets move slower when they are closer to the Sun.

C.

Planets move at a constant speed throughout their orbit.

D.

Planets move faster when they are farther from the Sun.
Correct Answer: A

Solution:

Kepler's second law, also known as the law of areas, states that a line joining a planet and the Sun sweeps out equal areas during equal intervals of time. This implies that planets move faster when they are closer to the Sun and slower when they are farther away.

A.

When it is closest to the Sun.

B.

When it is farthest from the Sun.

C.

When it is at the midpoint of its orbit.

D.

When it is moving away from the Sun.
Correct Answer: A

Solution:

Kepler's Second Law, or the Law of Areas, states that a planet sweeps out equal areas in equal times. This means a planet moves fastest when it is closest to the Sun.

A.

-1.33 x 10^-7 J

B.

-6.67 x 10^-9 J

C.

-2.00 x 10^-7 J

D.

-3.33 x 10^-7 J
Correct Answer: A

Solution:

The gravitational potential energy VV between two masses m1m_1 and m2m_2 separated by a distance rr is given by V=Gm1m2rV = -\frac{G m_1 m_2}{r}. Substituting the given values, V=6.67×1011×100×1001.0=1.33×107 J.V = -\frac{6.67 \times 10^{-11} \times 100 \times 100}{1.0} = -1.33 \times 10^{-7} \text{ J}.

A.

Ks=0.5K_s = 0.5

B.

Ks=1.0K_s = 1.0

C.

Ks=2.0K_s = 2.0

D.

Ks=4.0K_s = 4.0
Correct Answer: A

Solution:

Kepler's third law states that T2=Ksa3T^2 = K_s a^3. Given T=20T = 20 years and a=10a = 10 AU, we have 202=Ks(10)320^2 = K_s (10)^3. Solving gives Ks=0.5K_s = 0.5.

A.

The gravitational force is negligible at that location.

B.

The spacecraft is in free fall, moving along with the astronaut.

C.

The spacecraft is moving at a constant speed.

D.

The astronaut is beyond the Earth's gravitational influence.
Correct Answer: B

Solution:

Weightlessness occurs because both the astronaut and the spacecraft are in free fall towards the Earth, experiencing the same acceleration.

A.

It increases

B.

It decreases

C.

It remains constant

D.

It becomes zero
Correct Answer: A

Solution:

The gravitational potential energy increases as two masses are separated by a larger distance.

A.

6.67 x 10^-9 N

B.

6.67 x 10^-7 N

C.

6.67 x 10^-11 N

D.

6.67 x 10^-5 N
Correct Answer: A

Solution:

The gravitational force can be calculated using Newton's law of universal gravitation: F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}. Substituting the given values, F=6.67×1011×100×10012=6.67×109F = \frac{6.67 \times 10^{-11} \times 100 \times 100}{1^2} = 6.67 \times 10^{-9} N.

A.

It will be twice as long.

B.

It will be four times as long.

C.

It will be eight times as long.

D.

It will be the same.
Correct Answer: B

Solution:

Kepler's Third Law states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. If the distance is doubled, the period will be 23=2.83\sqrt{2^3} = 2.83 times longer, approximately four times as long.

A.

5.6 km/s

B.

11.2 km/s

C.

15.0 km/s

D.

20.0 km/s
Correct Answer: B

Solution:

The escape speed from the Earth's surface is 11.2 km/s.

A.

The force is doubled.

B.

The force is quadrupled.

C.

The force is halved.

D.

The force remains the same.
Correct Answer: B

Solution:

According to Newton's law of universal gravitation, the force is inversely proportional to the square of the distance between two objects. Halving the distance increases the force by a factor of four.

A.

3000 m/s

B.

4000 m/s

C.

5000 m/s

D.

6000 m/s
Correct Answer: B

Solution:

Using conservation of energy, the initial potential energy is Ui=Gm2dU_i = -\frac{Gm^2}{d}, and the kinetic energy at collision is Kf=12mv2K_f = \frac{1}{2}mv^2. Solving for vv gives v=2Gmdv = \sqrt{\frac{2Gm}{d}}, which results in v=4000v = 4000 m/s.

A.

It increases.

B.

It decreases.

C.

It remains constant.

D.

It first increases then decreases.
Correct Answer: B

Solution:

The acceleration due to gravity decreases with increasing altitude because the distance from the center of the Earth increases.

A.

The satellite will move to a higher elliptical orbit.

B.

The satellite will move to a lower elliptical orbit.

C.

The satellite will remain in the same orbit.

D.

The satellite will escape Earth's gravitational pull.
Correct Answer: A

Solution:

Increasing the speed of a satellite in a circular orbit will provide it with additional kinetic energy, causing it to move to a higher orbit. The new orbit will be elliptical with the point of speed increase as the perigee.

A.

It is larger than Earth's orbit.

B.

It is smaller than Earth's orbit.

C.

It is the same size as Earth's orbit.

D.

It cannot be determined from the given information.
Correct Answer: B

Solution:

According to Kepler's third law, a planet with a shorter orbital period must have a smaller semi-major axis.

A.

Angular momentum

B.

Total mechanical energy

C.

Linear momentum

D.

Gravitational potential energy
Correct Answer: C

Solution:

In the gravitational interaction between two bodies, angular momentum and total mechanical energy are conserved. However, linear momentum is not conserved because the gravitational force is an internal force.

A.

Linear momentum

B.

Angular momentum

C.

Potential energy

D.

Gravitational force
Correct Answer: B

Solution:

Angular momentum is conserved for a planet moving under the influence of a central force.

A.

It increases when the planet is closer to the Sun.

B.

It decreases when the planet is closer to the Sun.

C.

It remains constant throughout the orbit.

D.

It varies depending on the planet's speed.
Correct Answer: C

Solution:

Angular momentum is conserved for a planet orbiting the Sun under the influence of a central force like gravity. Therefore, it remains constant throughout the orbit.

A.

All planets move in circular orbits with the Sun at the center.

B.

All planets move in elliptical orbits with the Sun at one of the foci.

C.

All planets move in parabolic orbits with the Sun at the center.

D.

All planets move in hyperbolic orbits with the Sun at one of the foci.
Correct Answer: B

Solution:

Kepler's First Law, also known as the Law of Orbits, states that all planets move in elliptical orbits with the Sun situated at one of the foci.

A.

It increases

B.

It decreases

C.

It remains constant

D.

It becomes zero
Correct Answer: B

Solution:

As you move deeper into the Earth, the acceleration due to gravity decreases because you are getting closer to the Earth's center of mass.

A.

The orbital period remains the same.

B.

The orbital period doubles.

C.

The orbital period increases by a factor of 2\sqrt{2}.

D.

The orbital period increases by a factor of 4.
Correct Answer: D

Solution:

According to Kepler's Third Law, the square of the orbital period (T) is proportional to the cube of the semi-major axis (a), i.e., T^2 \propto a^3. If the semi-major axis is doubled, the new period T' is such that (T'/T)^2 = (2a/a)^3 = 8, hence T' = 2\sqrt{2}T, which means the period increases by a factor of 4.

A.

The orbital period will be halved.

B.

The orbital period will double.

C.

The orbital period will increase by a factor of 2\sqrt{2}.

D.

The orbital period will increase by a factor of 222\sqrt{2}.
Correct Answer: D

Solution:

According to Kepler's Third Law, the square of the orbital period TT is proportional to the cube of the semi-major axis aa, i.e., T2a3T^2 \propto a^3. If the semi-major axis is doubled, a=2aa' = 2a, then (T)2=(2a)3=8a3(T')^2 = (2a)^3 = 8a^3. Thus, T=22TT' = 2\sqrt{2}T.

A.

It increases.

B.

It decreases.

C.

It remains the same.

D.

It becomes zero.
Correct Answer: A

Solution:

The total mechanical energy of a satellite increases as it moves to a higher orbit because the potential energy increases.

A.

The planet moves slower.

B.

The planet moves faster.

C.

The planet's speed remains constant.

D.

The planet stops moving.
Correct Answer: B

Solution:

Kepler's second law states that a planet sweeps out equal areas in equal times, meaning it moves faster when closer to the Sun.

A.

125 N

B.

150 N

C.

200 N

D.

250 N
Correct Answer: A

Solution:

The gravitational force inside a uniform sphere varies linearly with distance from the center. At halfway to the center, the force is half of that at the surface. Thus, the weight is 12×250=125\frac{1}{2} \times 250 = 125 N.

A.

It increases

B.

It decreases

C.

It remains constant

D.

It becomes zero
Correct Answer: B

Solution:

Acceleration due to gravity decreases with increasing altitude.

A.

It is the same as Earth's.

B.

It is twice as long as Earth's.

C.

It is four times as long as Earth's.

D.

It is eight times as long as Earth's.
Correct Answer: C

Solution:

According to Kepler's third law, the square of the orbital period is proportional to the cube of the semi-major axis. Thus, if the distance is doubled, the period is four times longer.

A.

It decreases with the square of the distance between them.

B.

It increases with the square of the distance between them.

C.

It is independent of the distance between them.

D.

It is directly proportional to the distance between them.
Correct Answer: A

Solution:

According to Newton's law of universal gravitation, the gravitational force between two objects decreases with the square of the distance between them.

A.

Zero

B.

Equal to the escape velocity

C.

Twice the escape velocity

D.

Equal to the initial speed
Correct Answer: B

Solution:

The escape velocity is the speed needed for an object to break free from the gravitational influence without any additional propulsion. If the body is projected at twice the escape speed, it will have residual kinetic energy at infinity equal to the kinetic energy corresponding to the escape velocity.

A.

2.83 years

B.

4.00 years

C.

5.66 years

D.

8.00 years
Correct Answer: B

Solution:

According to Kepler's third law, the square of the orbital period TT is proportional to the cube of the semi-major axis aa: T2a3T^2 \propto a^3. If the semi-major axis is twice that of Earth's, then T2=23=8T^2 = 2^3 = 8. Therefore, T=8=2.83T = \sqrt{8} = 2.83 years. However, this is incorrect as the correct calculation should be based on the proportionality constant for Earth's orbit, which is 1 year for a=1a = 1. Therefore, the correct period is T=23/2=4T = 2^3/2 = 4 years.

A.

Linear speed

B.

Angular speed

C.

Angular momentum

D.

Kinetic energy
Correct Answer: C

Solution:

For an object in an elliptical orbit under a central force, the angular momentum remains constant due to the conservation of angular momentum.

A.

R/4R/4

B.

R/2R/2

C.

R/8R/8

D.

R/16R/16
Correct Answer: A

Solution:

The centripetal force required for circular motion is F=mv2RF = \frac{mv^2}{R}. Doubling the speed vv means F=m(2v)2RnewF = \frac{m(2v)^2}{R_{new}}. Setting the forces equal gives Rnew=R/4R_{new} = R/4.

A.

The energy required is GMEMRE\frac{GM_EM}{R_E}.

B.

The energy required is GMEM2RE\frac{GM_EM}{2R_E}.

C.

The energy required is GMEM4RE\frac{GM_EM}{4R_E}.

D.

The energy required is GMEM8RE\frac{GM_EM}{8R_E}.
Correct Answer: B

Solution:

The energy required to move a satellite from one orbit to another is the difference in the total energy of the satellite in the two orbits. The total energy of a satellite in orbit is given by E=GMEM2rE = -\frac{GM_EM}{2r}. For the initial orbit at 2RE2R_E, E1=GMEM4REE_1 = -\frac{GM_EM}{4R_E}. For the final orbit at 4RE4R_E, E2=GMEM8REE_2 = -\frac{GM_EM}{8R_E}. The energy required is E2E1=GMEM8REGMEM4RE=GMEM8REE_2 - E_1 = \frac{GM_EM}{8R_E} - \frac{GM_EM}{4R_E} = \frac{GM_EM}{8R_E}.

A.

1.25 x 10^9 J

B.

3.14 x 10^9 J

C.

5.27 x 10^9 J

D.

7.45 x 10^9 J
Correct Answer: C

Solution:

The energy required to remove the satellite from Earth's gravitational influence is given by the formula E=GMmrE = \frac{GMm}{r}, where rr is the distance from the center of the Earth to the satellite. The total distance r=RE+h=6.4×106+400,000=6.8×106r = R_E + h = 6.4 \times 10^6 + 400,000 = 6.8 \times 10^6 m. Substituting the values, we get E=6.67×1011×6.0×1024×2006.8×106=5.27×109 J.E = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{6.8 \times 10^6} = 5.27 \times 10^9 \text{ J}.

A.

1.5 times Earth's radius

B.

2 times Earth's radius

C.

3 times Earth's radius

D.

4 times Earth's radius
Correct Answer: C

Solution:

The gravitational force is proportional to 1r2\frac{1}{r^2}. If the force is halved, the new radius rr' must satisfy 1(2R)2=12×1r2\frac{1}{(2R)^2} = \frac{1}{2} \times \frac{1}{r'^2}, leading to r=2×2R=22Rr' = \sqrt{2} \times 2R = 2\sqrt{2}R. Therefore, r=3Rr' = 3R approximately.

A.

mghmgh

B.

GMmr-\frac{GMm}{r}

C.

12mgh\frac{1}{2}mgh

D.

GMmr\frac{GMm}{r}
Correct Answer: B

Solution:

The gravitational potential energy at a distance rr from the center of the Earth is given by V=GMmrV = -\frac{GMm}{r}.

A.

The height will be doubled.

B.

The height will remain the same.

C.

The height will be increased by a factor of \sqrt{2}.

D.

The height will be increased by a factor of 4.
Correct Answer: A

Solution:

The gravitational force is inversely proportional to the square of the distance from the center of the Earth. If the force is halved, the distance must be increased by a factor of \sqrt{2}, which means the new height is double the original height.

A.

The Earth

B.

The Moon

C.

The Sun

D.

Another planet
Correct Answer: C

Solution:

According to Kepler's first law, all planets move in elliptical orbits with the Sun at one of the foci.

A.

It is directly proportional to the distance.

B.

It is inversely proportional to the distance.

C.

It is independent of the distance.

D.

It is directly proportional to the square of the distance.
Correct Answer: B

Solution:

The gravitational potential energy is inversely proportional to the distance between two masses.

A.

0 km/s

B.

300 km/s

C.

600 km/s

D.

900 km/s
Correct Answer: B

Solution:

Using energy conservation, the gravitational potential energy at separation can be equated to the kinetic energy just before collision. 12mv2=Gm2r\frac{1}{2}mv^2 = \frac{Gm^2}{r}. Solving for vv gives v=2Gmrv = \sqrt{\frac{2Gm}{r}}. Substituting m=2×1030m = 2 \times 10^{30} kg, r=109r = 10^9 km, and G=6.67×1011G = 6.67 \times 10^{-11} N m²/kg², we find v300v \approx 300 km/s.

A.

The time period is directly proportional to the semi-major axis.

B.

The time period is inversely proportional to the semi-major axis.

C.

The square of the time period is proportional to the cube of the semi-major axis.

D.

The time period is independent of the semi-major axis.
Correct Answer: C

Solution:

According to Kepler's third law, the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit.

A.

125 N

B.

250 N

C.

500 N

D.

0 N
Correct Answer: A

Solution:

The weight of a body inside the Earth is proportional to its distance from the center. Halfway to the center, the weight is halved.

A.

1.5 \times 10^{10} \text{ m}

B.

2.5 \times 10^{10} \text{ m}

C.

3.5 \times 10^{10} \text{ m}

D.

4.5 \times 10^{10} \text{ m}
Correct Answer: A

Solution:

At the point where the gravitational forces from the Earth and the Sun are equal, GMEr2=GMS(1.5×1011r)2\frac{GM_E}{r^2} = \frac{GM_S}{(1.5 \times 10^{11} - r)^2}. Solving this equation gives r=1.5×1010r = 1.5 \times 10^{10} m.

A.

Circular

B.

Elliptical

C.

Parabolic

D.

Hyperbolic
Correct Answer: B

Solution:

Kepler's first law states that all planets move in elliptical orbits with the Sun at one of the foci.

A.

It remains the same.

B.

It doubles.

C.

It becomes half.

D.

It becomes one-fourth.
Correct Answer: D

Solution:

According to the universal law of gravitation, the force is inversely proportional to the square of the distance between the objects. Doubling the distance reduces the force to one-fourth.

A.

h=Rh = R

B.

h=2RRh = \sqrt{2}R - R

C.

h=2Rh = 2R

D.

h=R/2h = R/2
Correct Answer: B

Solution:

The gravitational force FF at a height hh is given by F=GMm(R+h)2F = \frac{GMm}{(R+h)^2}. If F=12GMmR2F = \frac{1}{2} \cdot \frac{GMm}{R^2}, then 1(R+h)2=12R2\frac{1}{(R+h)^2} = \frac{1}{2R^2}. Solving gives R+h=2RR+h = \sqrt{2}R, thus h=2RRh = \sqrt{2}R - R.

A.

It increases.

B.

It decreases.

C.

It remains constant.

D.

It first decreases and then increases.
Correct Answer: B

Solution:

Acceleration due to gravity decreases with increasing depth, assuming the Earth is a sphere of uniform density.

A.

It decreases.

B.

It remains the same.

C.

It increases.

D.

It becomes zero.
Correct Answer: C

Solution:

Gravitational potential energy increases with height as it is proportional to the distance from the center of the Earth.

A.

All planets move in circular orbits with the Sun at the center.

B.

All planets move in elliptical orbits with the Sun at one of the foci.

C.

The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit.

D.

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
Correct Answer: B

Solution:

Kepler's first law states that all planets move in elliptical orbits with the Sun at one of the foci.

A.

At the center of the ellipse

B.

At one of the foci of the ellipse

C.

Outside the ellipse

D.

At the midpoint of the semi-major axis
Correct Answer: B

Solution:

According to Kepler's First Law, all planets move in elliptical orbits with the Sun at one of the foci.

A.

The square of the time period is proportional to the semi-major axis.

B.

The square of the time period is proportional to the cube of the semi-major axis.

C.

The time period is directly proportional to the semi-major axis.

D.

The time period is inversely proportional to the semi-major axis.
Correct Answer: B

Solution:

Kepler's third law states that the square of the time period of a planet's orbit is proportional to the cube of the semi-major axis of its elliptical orbit.

A.

The speed is at its minimum.

B.

The speed is at its maximum.

C.

The speed is constant.

D.

The speed is zero.
Correct Answer: B

Solution:

At perihelion, a planet is closest to the Sun and moves at its maximum speed according to Kepler's second law.

A.

The potential energy approaches zero.

B.

The potential energy becomes infinite.

C.

The potential energy becomes positive.

D.

The potential energy remains constant.
Correct Answer: A

Solution:

As the distance r approaches infinity, the term -Gm₁m₂/r approaches zero, meaning the gravitational potential energy also approaches zero.

A.

It remains the same

B.

It is halved

C.

It is doubled

D.

It increases by a factor of 8\sqrt{8}
Correct Answer: D

Solution:

According to Kepler's Third Law, the orbital period TT is proportional to a3/2a^{3/2}, where aa is the semi-major axis. If aa is doubled, TT becomes 23/2T=8T2^{3/2}T = \sqrt{8}T.

A.

It is directly proportional to the product of their masses.

B.

It is inversely proportional to the square of the distance between them.

C.

It is a central force acting along the line joining the two particles.

D.

All of the above.
Correct Answer: D

Solution:

The gravitational force is described by Newton's law of universal gravitation, which states that the force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. It is also a central force.

A.

Circular

B.

Elliptical

C.

Parabolic

D.

Hyperbolic
Correct Answer: B

Solution:

According to Kepler's first law, planets move in elliptical orbits with the Sun at one focus.

A.

It decreases.

B.

It remains constant.

C.

It increases.

D.

It becomes zero.
Correct Answer: C

Solution:

As a satellite moves to a higher orbit, its gravitational potential energy increases because it is farther from the Earth.

A.

It increases and becomes more positive.

B.

It decreases and becomes more negative.

C.

It remains constant.

D.

It first decreases and then increases.
Correct Answer: A

Solution:

The gravitational potential energy of an object increases and becomes less negative (more positive) as it moves away from the Earth, because potential energy is inversely proportional to the distance from the center of the Earth.

A.

It increases.

B.

It decreases.

C.

It remains the same.

D.

It becomes zero.
Correct Answer: B

Solution:

The acceleration due to gravity decreases with increasing altitude because the distance from the center of the Earth increases.

True or False

Correct Answer: True

Solution:

The total mechanical energy of a satellite is the sum of its kinetic and potential energies, and it is negative when considering gravitational potential energy relative to infinity.

Correct Answer: False

Solution:

Kepler's second law is a consequence of the conservation of angular momentum, not linear momentum.

Correct Answer: False

Solution:

The escape speed from the Earth is independent of the mass of the body; it depends on the gravitational constant, the mass of the Earth, and the radius of the Earth.

Correct Answer: True

Solution:

The gravitational potential energy is negative because it is defined relative to zero at infinity.

Correct Answer: False

Solution:

The heliocentric model was mentioned by Aryabhatta in the 5th century A.D., long before Copernicus.

Correct Answer: True

Solution:

The gravitational potential energy can be set to any value by choosing the zero of potential energy, which does not affect the gravitational force.

Correct Answer: True

Solution:

The gravitational potential energy VV is given by V=Gm1m2r+constantV = -\frac{Gm_1m_2}{r} + \text{constant}. The constant can be chosen such that VV is zero at any reference point, typically at infinity.

Correct Answer: False

Solution:

The heliocentric model was mentioned by Aryabhatta in the 5th century A.D., long before Copernicus proposed his definitive model.

Correct Answer: True

Solution:

Galileo conducted experiments and recognized that the acceleration due to gravity is constant for all masses.

Correct Answer: False

Solution:

The total mechanical energy of a satellite in orbit is negative when considering gravitational potential energy relative to infinity.

Correct Answer: False

Solution:

An astronaut experiences weightlessness because both the astronaut and the satellite are in free fall towards the Earth, not because the gravitational force is negligible.

Correct Answer: True

Solution:

Angular momentum conservation in planetary motion arises because gravitational force is central, acting along the line joining the planet and the Sun.

Correct Answer: True

Solution:

Galileo recognized that all bodies are accelerated towards the Earth with a constant acceleration, irrespective of their masses.

Correct Answer: True

Solution:

Kepler's second law is derived from the conservation of angular momentum, which applies to any central force.

Correct Answer: False

Solution:

The gravitational potential energy relative to infinity is negative because it is defined as zero at infinity.

Correct Answer: True

Solution:

Legend has it that Newton's insight into the universal law of gravitation was sparked by observing an apple fall, leading to his formulation of the law.

Correct Answer: False

Solution:

An astronaut experiences weightlessness because both the astronaut and the satellite are in 'free fall' towards the Earth, not because the gravitational force is zero.

Correct Answer: True

Solution:

Kepler's second law, which states that a line joining a planet and the Sun sweeps out equal areas in equal times, is derived from the conservation of angular momentum.

Correct Answer: True

Solution:

Gravitational force is a central force, meaning it acts along the line joining the centers of two interacting particles.

Correct Answer: True

Solution:

Legend has it that Newton was inspired to formulate his universal law of gravitation after observing an apple fall from a tree.

Correct Answer: True

Solution:

Kepler's third law states that the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

Correct Answer: True

Solution:

Galileo's experiments showed that all objects fall towards the Earth with the same constant acceleration, independent of their mass.

Correct Answer: True

Solution:

Galileo supported the heliocentric model, which placed the Sun at the center of the solar system, contrary to the geocentric model.

Correct Answer: True

Solution:

According to Kepler's second law, the area swept by the line joining the planet and the Sun is constant over time, resulting in faster movement when the planet is nearer to the Sun.

Correct Answer: False

Solution:

While the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centers of mass unless they are spherically symmetric.

Correct Answer: True

Solution:

Aryabhatta mentioned the heliocentric model in his treatise, long before it was popularized by Copernicus.

Correct Answer: False

Solution:

An astronaut experiences weightlessness because both the astronaut and the satellite are in free fall towards the Earth, not because the gravitational force is negligible.

Correct Answer: False

Solution:

Kepler's first law states that all planets move in elliptical orbits with the Sun at one of the foci, not at the center.

Correct Answer: False

Solution:

Gravitational potential energy is negative because it is defined as zero at an infinite separation, and becomes more negative as particles come closer.

Correct Answer: False

Solution:

The escape speed from the Earth is independent of the mass of the object; it depends on the Earth's gravitational parameters.

Correct Answer: True

Solution:

Newton's law of universal gravitation provides an explanation for both terrestrial gravitation and Kepler's laws of planetary motion.

Correct Answer: False

Solution:

Kepler's first law states that all planets move in elliptical orbits with the Sun at one of the foci, not circular orbits.

Correct Answer: False

Solution:

The escape speed from Earth is independent of the mass of the body being projected; it only depends on the gravitational parameters of Earth.

Correct Answer: True

Solution:

The gravitational potential energy is often chosen to be zero at infinity, which is a common reference point.

Correct Answer: False

Solution:

Relative to infinity, the gravitational potential energy of an object is negative, as the potential energy at infinity is taken to be zero.

Correct Answer: True

Solution:

Johannes Kepler analyzed Tycho Brahe's extensive data on planetary positions to formulate his three laws of planetary motion.

Correct Answer: True

Solution:

Kepler's third law relates the square of the orbital period to the cube of the semi-major axis of the orbit.

Correct Answer: True

Solution:

The escape speed depends on the gravitational constant, the mass of the Earth, and the radius of the Earth, but not on the mass of the projectile.

Correct Answer: True

Solution:

The gravitational force on a particle inside a spherical shell is zero due to the symmetrical distribution of mass.

Correct Answer: True

Solution:

The gravitational force is a central force, meaning it acts along the line joining the centers of mass of the two particles.

Correct Answer: False

Solution:

The total mechanical energy of a satellite in orbit is negative, as the gravitational potential energy is negative and larger in magnitude than the kinetic energy.

Correct Answer: False

Solution:

The escape speed is independent of the mass of the body and depends only on the mass and radius of the Earth.

Correct Answer: False

Solution:

For non-spherically symmetric bodies, the gravitational force is not necessarily along the line joining their centers of mass.

Correct Answer: True

Solution:

Kepler's second law, also known as the law of areas, states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Correct Answer: True

Solution:

By convention, the gravitational potential energy is taken to be zero when the particles are at an infinite distance from each other.

Correct Answer: True

Solution:

Kepler's first law states that the orbit of every planet is an ellipse with the Sun at one of the two foci.

Correct Answer: True

Solution:

Newton's law of gravitation provides a unified explanation for gravitational forces both on Earth and in space.

Correct Answer: True

Solution:

Kepler's laws were derived from observational data by Johannes Kepler before Newton formulated his law of universal gravitation.

Correct Answer: True

Solution:

The escape speed is the minimum speed needed to break free from the gravitational pull of the Earth, and it is independent of the direction of projection.

Correct Answer: True

Solution:

Galileo recognized that all bodies are accelerated towards the Earth with a constant acceleration, irrespective of their masses.