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 ${{v}_{0}}=\sqrt{2gh}$ relative to the earth. Obtain the same formula in terms of the reference frame “falling” to the earth with a constant velocity ${{v}_{0}}$ .

Answer 1

We have to consider a moving frame of reference. Therefore we need to consider that earth has a velocity as well. We need to use the equation of motion. We should apply conservation of energy equation, to find the final velocity of the earth and the initial velocity of the stone.
Formula used:
Conservation of energy: $\dfrac{1}{2}mv_{0}^{2}+\dfrac{1}{2}Mv_{0}^{2}+mgh=\dfrac{1}{2}M{{({{v}_{0}}+\dfrac{m}{M}{{v}_{0}})}^{2}}$

Complete answer:
Here we have to consider a non-inertial frame of reference so we take into account the kinetic energy of the earth and the acceleration of the earth. These energies need to be applied in the conservation of energy equation. In the non- inertial frame of reference, the earth and the stone are moving at the same velocity ${{v}_{0}}$.
We have to use the equation of motion to find final velocity of the stone:
$0={{v}_{0}}-gt$ (Since final velocity is zero)
Final velocity of the earth
${{v}_{0}}-\dfrac{m}{M}gt$ (Where M is mass of earth)
Let us apply conservation of energy
$\dfrac{1}{2}mv_{0}^{2}+\dfrac{1}{2}Mv_{0}^{2}+mgh=\dfrac{1}{2}M{{({{v}_{0}}+\dfrac{m}{M}{{v}_{0}})}^{2}}$
On simplifying the above equation, we get
$mgh=\dfrac{1}{2}v_{0}^{2}(m+\dfrac{{{m}^{2}}}{M})$
In the above equation, we can neglect $\dfrac{{{m}^{2}}}{M}$ as $M\gg {{m}^{2}}$
$\begin{aligned} & \Rightarrow v_{0}^{2}=2gh \\\ & \Rightarrow {{v}_{0}}=\sqrt{2gh} \\\ \end{aligned}$
$\therefore {{v}_{0}}=\sqrt{2gh}$

Note:
In the rest frame of earth, earth’s acceleration is of the order $\dfrac{m}{M}$ and is neglected. In the non-inertial frame, the acceleration can be used as it is if the order one and that is large. When we are comparing the mass of a stone to the mass of earth then the stone is very insignificant. So the fraction $\dfrac{m}{M}$ tends to zero and that’s the reason it was neglected by us here.