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Physics Part-I

Gravitation

As you have learnt in the text, a geostationary satellite orbi

Physics Part-I

Class 11

NCERT

Chapter 1: Physical World

16 questions

Chapter 2: Units and Measurements

51 questions

Chapter 3: Motion in Straight Line

50 questions

Chapter 4: Motion in a Plane

43 questions

Chapter 5: Laws of motion

52 questions

Chapter 6: Work, Energy and Power

41 questions

Chapter 7: System of Particles and Rotational Motion

45 questions

Chapter 8: Gravitation

33 questions

Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth =

6.0 \times 1 0^{24} k g

, radius = 6400 km.

A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun =

2 \times 1 0^{30} k g

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As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = $6.0 \times 1 0^{24} k g$ , radius = 6400 km.

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Physics Part-I (NCERT)

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Oliver Discussed

6.0 \times 1 0^{24} k g

, radius = 6400 km.

11 mins ago

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11 mins ago

Text solutionVerified

Mass of the Earth,

M = 6.0 \times 1 0^{24} k g

Radius of the Earth,

R = 6400 km = 6.4 \times 1 0^{6} m

Height of a geostationary satellite from the surface of the Earth,

h = 36000 km = 3.6 \times 1 0^{7} m

Gravitational potential due to Earth's gravity at height h

V = - \frac{GM}{R + h}

= - \frac{6.67 \times 1 0 ^{- 11} \times 6 \times 1 0 ^{24}}{3.6 \times 1 0 ^{7} + 0.64 \times 1 0 ^{7}}

= - 9.4 \times 1 0^{6} J / k g

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6.0 \times 1 0^{24} k g

, radius = 6400 km.

2 \times 1 0^{30} k g

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Question Text	As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = $6.0 \times 1 0^{24} k g$ , radius = 6400 km.
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	10

Text solutionVerified

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