Question

A satellite is moving with speed v in a circular orbit about the earth. An object of mass 2m is ejected from the satellite such that it just escapes from the gravitational pull of earth. At the time of ejection the kinetic energy of object is

Answer 1

The answer is 2)mv^2.

EXPLANATION:

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