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Class 11 Science (PCM)

Physics Part 1

Gravitation

Gravitation

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Gravitational Constant Measurement
Accuracy: 0%
Total Attempted4
Total Correct0
Total Incorrect4
Difficulty Level:
High
Gravitational Potential
Accuracy: 0%
Total Attempted1
Total Correct0
Total Incorrect1
Difficulty Level:
High
Weightlessness and Free Fall in Satellites
Accuracy: 20%
Total Attempted5
Total Correct1
Total Incorrect4
Difficulty Level:
High
Acceleration Due to Gravity
Accuracy: 29%
Total Attempted7
Total Correct2
Total Incorrect5
Difficulty Level:
High
Earth Satellites and Orbital Motion
Accuracy: 38%
Total Attempted8
Total Correct3
Total Incorrect5
Difficulty Level:
High
Energy of an Orbiting Satellite
Accuracy: 43%
Total Attempted7
Total Correct3
Total Incorrect4
Difficulty Level:
High
Gravitational Potential Energy
Accuracy: 43%
Total Attempted7
Total Correct3
Total Incorrect4
Difficulty Level:
High
Universal Law of Gravitation
Accuracy: 50%
Total Attempted6
Total Correct3
Total Incorrect3
Difficulty Level:
Medium
Gravitational Force in Spherical Shells
Accuracy: 55%
Total Attempted11
Total Correct6
Total Incorrect5
Difficulty Level:
Medium
Escape Speed
Accuracy: 67%
Total Attempted6
Total Correct4
Total Incorrect2
Difficulty Level:
Medium
Kepler's Laws of Planetary Motion
Accuracy: 100%
Total Attempted9
Total Correct9
Total Incorrect0
Difficulty Level:
Low
Superposition of Gravitational Forces
Accuracy: 100%
Total Attempted1
Total Correct1
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand Kepler's Laws of Planetary Motion, including the elliptical orbits of planets, the equal area law, and the relationship between orbital period and semi-major axis.
  • Calculate gravitational forces using Newton's Universal Law of Gravitation, which states that every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
  • Analyze the concept of Gravitational Potential Energy, defined as the work done in bringing a particle from infinity to a point in space.
  • Determine the Escape Speed required for an object to break free from the gravitational attraction of a celestial body without further propulsion.
  • Evaluate the variation of Acceleration Due to Gravity with altitude and depth, and understand its implications for objects near the Earth's surface.
  • Compute the Energy of an Orbiting Satellite, which includes both kinetic and potential energy, and understand its implications for satellite motion.
  • Explore the methods for Gravitational Constant Measurement, such as Cavendish's experiment, to determine the value of the gravitational constant GG.
  • Examine the Gravitational Force in Spherical Shells, including the zero force inside a shell and the force acting as if mass is concentrated at the center for external points.
  • Apply the Superposition of Gravitational Forces to calculate the resultant gravitational force due to multiple masses using vector addition and symmetry-based cancellation.
  • Understand the concept of Gravitational Potential as potential energy per unit mass, its relation to gravitational potential energy, and the superposition of potentials.
  • Analyze the motion of Earth Satellites and Orbital Motion, including circular satellite orbits, orbital speed, time period, and the application of Kepler’s law to satellites.
  • Explain the phenomenon of Weightlessness and Free Fall in Satellites, where astronauts experience weightlessness because both the satellite and the astronaut are in free fall.

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Revision Notes & Summary

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Chapter Notes

Kepler's Laws of Planetary Motion

  • Law of Orbits: All planets move in elliptical orbits with the Sun at one of the foci.
  • Law of Areas: The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
  • Law of Periods: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

Universal Law of Gravitation

  • Newton's Law: Every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
  • Formula: F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

Gravitational Potential Energy

  • Definition: The potential energy due to gravitational forces, defined as the work done in bringing a particle from infinity to a point in space.
  • Formula: V=−Gm1m2rV = -G \frac{m_1 m_2}{r}

Escape Speed

  • Definition: The minimum speed required for an object to break free from the gravitational attraction of a celestial body without further propulsion.
  • Formula: ve=2GMRv_e = \sqrt{\frac{2GM}{R}}

Acceleration Due to Gravity

  • Surface of Earth: g=GMR2g = \frac{GM}{R^2}
  • Variation with Altitude: g(h)=g(0)(1−2hR)g(h) = g(0) \left(1 - \frac{2h}{R}\right)
  • Variation with Depth: g(d)=g(0)(1−dR)g(d) = g(0) \left(1 - \frac{d}{R}\right)

Energy of an Orbiting Satellite

  • Total Energy: The sum of kinetic and potential energy, negative for bound systems.
  • Formula: E=−GMm2aE = -\frac{GMm}{2a}

Gravitational Constant Measurement

  • Cavendish Experiment: Used to measure the gravitational constant GG.

Gravitational Force in Spherical Shells

  • Outside a Shell: Acts as if the mass is concentrated at the center.
  • Inside a Shell: The gravitational force is zero.

Superposition of Gravitational Forces

  • Principle: Resultant force due to multiple masses is the vector sum of individual forces.

Gravitational Potential

  • Definition: Potential energy per unit mass.
  • Relation: V=−GMrV = -\frac{GM}{r}

Earth Satellites and Orbital Motion

  • Orbital Speed: v=GMR+hv = \sqrt{\frac{GM}{R+h}}
  • Time Period: T=2π(R+h)3GMT = 2\pi \sqrt{\frac{(R+h)^3}{GM}}

Weightlessness and Free Fall in Satellites

  • Explanation: Astronauts experience weightlessness because both they and the satellite are in free fall towards Earth.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

The energy required is ΔE=GMEm8RE\Delta E = \frac{G M_E m}{8R_E}.

B.

The energy required is ΔE=GMEm4RE\Delta E = \frac{G M_E m}{4R_E}.

C.

The energy required is ΔE=GMEm2RE\Delta E = \frac{G M_E m}{2R_E}.

D.

The energy required is ΔE=GMEmRE\Delta E = \frac{G M_E m}{R_E}.
Correct Answer: A

Solution:

The change in energy is calculated using the formula for gravitational potential energy, and the difference between the two orbits is ΔE=GMEm8RE\Delta E = \frac{G M_E m}{8R_E}.

Chapter Concept:

Gravitational Potential Energy

A.

−GMEmr-\frac{G M_E m}{r}

B.

GMEmr\frac{G M_E m}{r}

C.

−GMEmr2-\frac{G M_E m}{r^2}

D.

GMEmr2\frac{G M_E m}{r^2}
Correct Answer: A

Solution:

Gravitational potential energy is given by V=−GMEmrV = -\frac{G M_E m}{r}, where GG is the gravitational constant, MEM_E is Earth's mass, and rr is the distance from the center.

Chapter Concept:

Universal Law of Gravitation

A.

−4Gm2l-\frac{4Gm^2}{l}

B.

−22Gm2l-\frac{2\sqrt{2}Gm^2}{l}

C.

−6Gm2l-\frac{6Gm^2}{l}

D.

−8Gm2l-\frac{8Gm^2}{l}
Correct Answer: C

Solution:

The potential energy between each pair of masses is −Gm2r-\frac{Gm^2}{r}, where rr is the distance between them. For four masses at the vertices of a square, there are four pairs at distance ll and two diagonal pairs at distance 2l\sqrt{2}l. Therefore, the total potential energy is 4(−Gm2l)+2(−Gm22l)=−6Gm2l4\left(-\frac{Gm^2}{l}\right) + 2\left(-\frac{Gm^2}{\sqrt{2}l}\right) = -\frac{6Gm^2}{l}.

Chapter Concept:

Superposition of Gravitational Forces

A.

It increases linearly with depth.

B.

It decreases linearly with depth.

C.

It remains constant.

D.

It becomes zero.
Correct Answer: B

Solution:

The acceleration due to gravity decreases with depth, as given by g(d)=g(0)(1−dRE)g(d) = g(0)(1 - \frac{d}{R_E}).

Chapter Concept:

Acceleration Due to Gravity

A.

It increases.

B.

It decreases.

C.

It remains constant.

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.

Chapter Concept:

Acceleration Due to Gravity

A.

−5.88×109 J-5.88 \times 10^9 \text{ J}

B.

−7.84×109 J-7.84 \times 10^9 \text{ J}

C.

−8.96×109 J-8.96 \times 10^9 \text{ J}

D.

−9.12×109 J-9.12 \times 10^9 \text{ J}
Correct Answer: B

Solution:

The gravitational potential energy is given by U=−GMEmRE+hU = -\frac{G M_E m}{R_E + h}. Substituting the given values, U=−6.67×10−11×6.0×1024×2006.4×106+400×103=−7.84×109 JU = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{6.4 \times 10^6 + 400 \times 10^3} = -7.84 \times 10^9 \text{ J}.

Chapter Concept:

Gravitational Potential Energy

A.

U=−GMEmRE+hU = -\frac{G M_E m}{R_E + h}

B.

U=GMEmRE+hU = \frac{G M_E m}{R_E + h}

C.

U=−GMEmREU = -\frac{G M_E m}{R_E}

D.

U=GMEmREU = \frac{G M_E m}{R_E}
Correct Answer: A

Solution:

The gravitational potential energy of a satellite at a height hh is given by U=−GMEmRE+hU = -\frac{G M_E m}{R_E + h}, where MEM_E is the Earth's mass and RER_E is the Earth's radius.

Chapter Concept:

Gravitational Potential

A.

V2=GMERE+hV^2 = \frac{GM_E}{R_E + h}

B.

V2=GME(RE+h)V^2 = \frac{G}{M_E (R_E + h)}

C.

V2=GME(RE+h)2V^2 = \frac{G M_E (R_E + h)}{2}

D.

V2=2GMERE+hV^2 = \frac{2 G M_E}{R_E + h}
Correct Answer: A

Solution:

The centripetal force required for the satellite's orbit is provided by the gravitational force, leading to the equation V2=GMERE+hV^2 = \frac{GM_E}{R_E + h}.

Chapter Concept:

Superposition of Gravitational Forces

A.

RER_E

B.

0.5RE0.5R_E

C.

1.5RE1.5R_E

D.

2RE2R_E
Correct Answer: A

Solution:

The gravitational force decreases with the square of the distance from the center of the Earth. If the force is half, the distance must be 2RE\sqrt{2}R_E, which means the height above the surface is 2RE−RE≈RE\sqrt{2}R_E - R_E \approx R_E.

Chapter Concept:

Escape Speed

A.

Kepler's First Law

B.

Kepler's Second Law

C.

Kepler's Third Law

D.

Newton's Law of Universal Gravitation
Correct Answer: B

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.

Chapter Concept:

Kepler's Laws of Planetary Motion

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Gravitation

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True or False

Correct Answer: True

Solution:

As altitude increases, the distance from the center of the Earth increases, leading to a decrease in gravitational force and thus a decrease in acceleration due to gravity.

Chapter Concept :

Acceleration Due to Gravity

Correct Answer: True

Solution:

According to the shell theorem, the gravitational force on a particle inside a uniform spherical shell is zero because the forces exerted by the shell cancel out.

Chapter Concept :

Universal Law of Gravitation

Correct Answer: True

Solution:

According to the principle of superposition and the properties of gravitational forces, the net gravitational force inside a uniform spherical shell is zero due to the symmetry of the shell.

Chapter Concept :

Superposition of Gravitational Forces

Correct Answer: True

Solution:

As depth increases below the Earth's surface, the acceleration due to gravity decreases because the effective mass attracting the object decreases.

Chapter Concept :

Acceleration Due to Gravity

Correct Answer: True

Solution:

Gravitational potential energy is indeed defined as the potential energy of a particle of unit mass at a point, as discussed in the context of gravitational forces.

Chapter Concept :

Gravitational Potential

Correct Answer: True

Solution:

The total mechanical energy of a satellite in orbit is negative because the potential energy is negative and twice the magnitude of the positive kinetic energy.

Chapter Concept :

Energy of an Orbiting Satellite

Correct Answer: True

Solution:

According to the superposition principle, the resultant gravitational force acting on a point mass due to multiple other masses is the vector sum of the individual forces exerted by each mass. This principle is fundamental in calculating gravitational forces in systems with multiple bodies.

Chapter Concept :

Superposition of Gravitational Forces

Correct Answer: True

Solution:

Escape speed is defined as the minimum speed needed for an object to escape from the gravitational influence of a celestial body without any additional propulsion.

Chapter Concept :

Escape Speed

Correct Answer: True

Solution:

Gravitational potential energy is indeed defined as the work done in bringing a particle from infinity to a point in space, as it is the energy stored due to gravitational forces.

Chapter Concept :

Gravitational Potential Energy

Correct Answer: True

Solution:

For satellites very close to the Earth's surface, the time period T0T_0 can be calculated using the formula T0=2πREgT_0 = 2\pi \sqrt{\frac{R_E}{g}}. Substituting g=9.8 m/s2g = 9.8 \text{ m/s}^2 and RE=6400 kmR_E = 6400 \text{ km}, we find that T0≈85 minutesT_0 \approx 85 \text{ minutes}.

Chapter Concept :

Earth Satellites and Orbital Motion

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