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A satellite is moving with constant speed v in a circular orbit around the earth.an object of mass m is ejected from the gravitational pull of the earth.at

the time of ejection,the kinetic energy of object is
1) 1/2 mv^2. 2)mv^2
3)3/2 mv^2. 4)2mv^2



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The answer is 2)mv^2.
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

2 5 2
hope it helps..
plss mark it as brainliest.....
Nice ans
The correct answers is option 2
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