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Grade 12IIT JEE Entrance Exam

A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is -

Sumanth Mede

7 Years agoGrade 12

1 Answer

Arun

7 Years ago

For a spherically symmetric body (earth), the escape velocity at a given distance is calculated by the formula Ve = √(2GM/R)

we need the orbital height, which we can get from:
Satellite motion, circular
V = √(GM/R)
V = velocity in m/s
G = 6.673e-11 Nm²/kg²
M is mass of central body in kg
R is radius of orbit in m

V² = GM/R
R = GM/V²

plugging that into the escape velocity equation, we get
Ve = √(2GM/(GM/V²)) = V√2

Kinetic Energy in J if m is in kg and v is in m/s
KE = ½mv² = ½m(V√2)² = mV²

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