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Question

A satellite of massm is launched vertically upward with an initial speed u from the surface of the earth. After it reaches heightR(R= radius of earth), it ejects a rocket of mass m10 so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G= gravitational constant;M is the mass of earth)


A

5mu2-119200GMR

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B

m20u-2GM3R2

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C

3m8u+5GM6R2

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D

m20u2+113200GMR

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Solution

The correct option is A

5mu2-119200GMR


Step 1. Given data :

Satellite mass =m

Speed =u

Radius of earth =R

Rocket mass =m10

Gravitational constant =G

Mass of earth =M

Step 2. Find the value of V :

Applying energy conservation

Ki+Ui=Kf+Uf

Where, Ki=initial kinetic energy

Ui=initial potential energy

Kf=final kinetic energy

Uf=final potential energy

12mu2+-GMmR=12mv2-GMm2R

v=u2-GMR …(1)

By momentum conservation, we have

m1u1+m2u2=m1v1+m2v2

Where,

m1is mass of the first body

m2 is the mass of the second body

u1 and u2 are the initial velocities and

v1 and v2 are the final velocities.

Let the rocket has velocity Vr​ in radial direction and VT tangential direction.

Applying momentum conservation in tangential direction.

m10vT=9m10GM2R ….(2)

and m10vr=mv

m10vr=mu2-GMR …..(3)

Step 3. Find the kinetic energy of the object :

kinetic energy of rocket =12×m10(VT2+Vr2)

=m20(81GM2R+100u2-100GMR)=m20100u2-119GM2R

=5mu2-119200GMR

Hence, (A) is the correct option.


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