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Physics Part-I

Gravitation

A star 2.5 times the mass of the sun and collapsed to a size o

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

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

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun =

2 \times 1 0^{30} k g

; mass of mars =

6.4 \times 1 0^{23} k g

; Radius of mars = 3395 km; Radius of the orbit of mars =

2.28 \times 1 0^{8} km

G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}

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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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2 \times 1 0^{30} k g

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Text solutionVerified

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

$Gravitational force, f_{g} = - GM m / R^{2}$

where

$Mass of the star, M = 2.5 \times 2 \times 1 0^{30} = 5 \times 1 0^{30} k g$

$Mass of the body : m$

$Radius of the star : R = 12 km = 1.2 \times 1 0^{4} m$

$f_{g} = \frac{6.67 \times 1 0 ^{- 11} \times 5 \times 1 0 ^{30}}{( 1.2 \times 1 0 ^{4} ) ^{2}} m$

$= 2.31 \times 1 0^{12} m N$

$Centrifugal force, f_{c} = m R ω^{2}$

$= m \times 4 π^{2} f^{2} R$

$= m \times 4 \times 3.1 4^{2} \times 1. 2^{2} \times 1.2 \times 1 0^{4}$

$= 6.82 \times 1 0^{5} m N$

Since $f_{g} > f_{c}$ , the body will remain stuck to the surface of the star.

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Practice more questions from Physics Part-I (NCERT)

6.0 \times 1 0^{24} k g

, radius = 6400 km.

2 \times 1 0^{30} k g

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun =

2 \times 1 0^{30} k g

; mass of mars =

6.4 \times 1 0^{23} k g

; Radius of mars = 3395 km; Radius of the orbit of mars =

2.28 \times 1 0^{8} km

G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}

View all

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Question Text	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 = $2 \times 1 0^{30} k g$ ).
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	28

Text solutionVerified

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