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What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

  • 5GmM/6R

  • 2GmM/3R

  • GmM/2R

  • GmM/2R


A.

5GmM/6R

From conservation of energy,
Total energy at the planet = Total energy at the altitude

In its orbit the necessary centripetal force provided by gravitational force.
∴ 

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Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

  • -4Gm/r

  • -6Gm/r

  • -9Gm/r

  • -9Gm/r


C.

-9Gm/r

GM over straight x squared space equals space fraction numerator straight G left parenthesis 4 straight m right parenthesis over denominator left parenthesis straight r minus straight x squared right parenthesis end fraction
1 over straight x space equals space fraction numerator 2 over denominator straight r minus straight x end fraction
straight r minus straight x space equals space 2 straight x
3 straight x space equals space straight r over 3


= -3Gm/r - 6Gm/r = -9Gm/r


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The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)


B.

straight g space equals space GMx over straight R cubed space inside space the space Earth space
straight g equals space GM over straight r squared space outside space the space Earth
Where M is mass of earth



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The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g'and 'R'(radius of earth) are 10 m/s2 and 6400km respectively. The required energy for this work will be;

  • 6.4 x1011 J

  • 6.4 x108 J

  • 6.4 x109 J

  • 6.4 x109 J


D.

6.4 x109 J

Potential energy on the earth's surface is -mgR while in free space, it is zero. So, to free the spaceship minimum required energy is
E =mgR
 = 103 x 10 x 6400 x 103 J
 = 6.4 x 1010 J

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The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is

  • 2R

  • fraction numerator straight R over denominator square root of 2 end fraction
  • R/2

  • R/2


A.

2R

fraction numerator GM over denominator 9 straight R squared end fraction space equals space fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction
rightwards double arrow space 3 straight R space equals space straight R plus straight h
rightwards double arrow space straight h equals space 2 straight R
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