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Question

A stone falls down without initial velocity from a height h onto the Earth's surface. The air drag assumed to be negligible, the stone hits the ground with velocity v0=2gh relative to the Earth. Obtain the same formula in terms of the reference frame "falling" to the Earth with a constant velocity v0.

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Solution

In a frame moving relative to the earth, one has to include the kinetic energy of the earth as well as earth's acceleration to be able to apply conservation of energy to the problem. In a reference frame failing to the earth with velocity v0, the stone is initially going up with velocity v0 and so is the earth. The final velocity of the stone is 0=v0gt and that of the earth is v0+mMgt (M is the mass of the earth), from Newton's third law, where t= time of fall. From conservation of energy
12mv20+12Mv20+mgh=12M(v0+mMv0)2
Hence 12v20(m+m2M)=mgh
Neglecting mM in comparison with 1, we get
v20=2gh or v0=2gh
The point is this in earth's rest frame the effect of earth's acceleration is of order mM and can be neglected but in a frame moving with respect to the earth the effect of earth's acceleration must be kept because it is of order one (i.e. large).

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