# A ball floats on the surface of water in a container exposed to atmosphere. Will the ball remain immersed at its initial depth or will it sink or rise some what if the container is shifted to moon?

Feb 13, 2018

#### Explanation:

If the container continues to be exposed to the atmosphere (of which there is none, or very little on the moon) the water will quickly boil off. It is possible there might be some ice remaining. It would be hard to say if the ball was floating or not during the short boiling period.

If the container is in a pressurized vehicle (or settlement) on the moon, the ball will still float. The change in gravity will not change the buoyancy equation. The local acceleration due to gravity is part of the weight/volume measurement of both the ball and the water, so it cancels out.

I hope this helps,
Steve

Feb 15, 2018

Let $m$ be mass of ball. ${\rho}_{w}$ density of water, ${\rho}_{a}$ density of air, ${v}_{w}$, volume of ball in water, ${v}_{a}$ volume of ball in air.
Let us write the complete expression for equilibrium on earth. We have

Weight of Ball $=$ Total upthrust
$m g = {v}_{w} {\rho}_{w} g + {v}_{a} {\rho}_{a} g$
$\implies m = {v}_{w} {\rho}_{w} + {v}_{a} {\rho}_{a}$ ..........(1)

We see that mass expression (1) is independent of gravity.

Now we know that moon's atmosphere is of infinitesimal amount as compared to Earth's atmosphere. Practically it can be taken as non-existent. As such equation (1) becomes

$m = {v}_{w}^{'} {\rho}_{w}$ .......(2)
where ${v}_{w}^{'}$ is volume of ball in water on moon.

On comparison of (1) and (2) we find that v_w^'>v_w^.
This implies that the ball will sink with reference to initial depth at earth's surface.

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Note: I think when the ball-container system is transferred from earth to moon, water will be sucked into moon's depleted atmosphere and the ball will drop down to the bottom of empty container.