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Gravitation

More Topic from Physics

Question 1
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A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between
(you may take G = 6 . 67× 10-11 Nm2/ kg2)

  • 13.34 x 10-10 J

  • 3.33 x 10-10 J

  • 6.67 x10-9 J

  • 6.67 x10-10 J

Solution

D.

6.67 x10-10 J

increment straight W space equals space straight v subscript straight f minus straight v subscript straight i space equals space 0 minus open square brackets fraction numerator negative GMm over denominator straight R end fraction close square brackets straight w space equals space fraction numerator space 6.67 space straight x space 10 to the power of negative 11 end exponent space space straight x 1000 over denominator.1 end fraction space straight x 10 over 1000 space equals space 6.67 space straight x space 10 to the power of negative 10 space end exponent straight J
Question 2
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A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 kms−1, the escape velocity from the

  • 1.1 kms−1

  • 11 kms−1

  • 110 kms−1

  • 0.11 kms−1

Solution

C.

110 kms−1

straight V subscript esc space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space equals space square root of fraction numerator 2 straight G space straight x space 10 space straight M over denominator straight R divided by 10 end fraction end root space space equals space 10 space straight x space 11 space equals space 110 space km divided by straight s
Question 3
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A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)

  • square root of 2 gR end root
  • square root of gR
  • square root of gR divided by 2 end root
  • square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

Solution

D.

square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of earth R, i.e h <<R
Orbit velocity of a satellite
straight v equals space square root of fraction numerator GM over denominator straight R plus straight h end fraction end root space equals space square root of GM over straight R end root space left parenthesis as space straight h less than less than straight R right parenthesis Velocity space required space to space escape 1 half space mv squared space equals space fraction numerator GMn over denominator straight R plus straight h end fraction straight v apostrophe space equals space square root of fraction numerator 2 GM over denominator straight R plus straight h end fraction end root space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space left parenthesis straight h less than less than straight R right parenthesis
therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.
straight v apostrophe minus straight v space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space minus square root of GM over straight R end root space equals space square root of 2 gR end root minus square root of gR space equals space square root of gR space left parenthesis square root of 2 minus 1 right parenthesis

Question 4
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A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

  • fraction numerator gR over denominator straight R minus straight x end fraction

  • gx

  • fraction numerator gR squared over denominator straight R space plus straight x end fraction
  • open parentheses fraction numerator gR squared over denominator straight R space plus straight x end fraction close parentheses to the power of 1 divided by 2 end exponent

Solution

D.

open parentheses fraction numerator gR squared over denominator straight R space plus straight x end fraction close parentheses to the power of 1 divided by 2 end exponent

The gravitational force exerted on satellite at a height x is
straight F subscript straight G space equals space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction
where Me = mass of earth Since, gravitational force provides the necessary centripetal force, so,
fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction space equals space fraction numerator mv subscript 0 superscript 2 over denominator left parenthesis straight R plus straight x right parenthesis end fraction rightwards double arrow space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2 rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2 rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2 space open parentheses because space straight g space equals space fraction numerator begin display style GM subscript straight e end style over denominator straight R end fraction close parentheses straight v subscript straight o space equals space square root of open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets end root space equals space open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets to the power of 1 divided by 2 end exponent

Question 5
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A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected?

  • moment of inertia

  • angular momentum

  • angular velocity

  • rotational kinetic energy

Solution

B.

angular momentum

In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.

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