Speed of falling objects / with different altitudes

Question

I have a question about gravity

So as we know gravity is significantly less on high mountains or tall buildings and increases as we lose height

☕ My question is that if we have 2 objects with the same volume and mass, 1 falls on a mountain and one falls on a beach ( note that they are released from the same height based on the surface they shall fall on ) Will they both reach the ground at the same time or will their speeds be different?

Answer 1

Gravity on the top of Mount Everest is approximately the same as gravity at sea level, with difference lower than $1\%$, usually negligible if you're doing such a kind of experiment.

Anyway, the answer is yes, the ball at the sea level takes less time than the ball on the top of the mount Everest, because the gravity is stronger and because the motion is governed by the same equation

$y(t) = h - \dfrac{1}{2} g t^2$,

where $g$ is the local intensity of the gravity field (approximately constant in the domain of your experiment), so that the time to reach the ground, i.e. when $y(t_G)=0$, is

$t_G = \sqrt{\dfrac{2 h}{g}}$.

Since $g^{sea} > g^{Eve}$, you get

$t^{sea}_G = \sqrt{\dfrac{2 h}{g^{sea}}} < \sqrt{\dfrac{2 h}{g^{Eve}}} = t^{Eve}_G$.

Answer 2

At different points on Earth's surface, the free fall acceleration ranges from 9.764 to 9.834 $m/s^2$.

$9.834 / 9.764 \approx 1.007 $

https://en.wikipedia.org/wiki/Gravitational_acceleration

$\text{speed}= g \space \text{time}$

$\text{distance} = \frac{1}{2} g \space \text{time}^2 $

$\text{g}= \text{acceleration of gravity}$

$g_\text{low} = 1.007 \space g_{\text{high}} $

$\frac{\text{speed}_\text{low}}{\text{speed}_\text{high}} = 1.007$

$\frac{\text{distance}_\text{low}}{\text{distance}_\text{high}} = 1.007$