Interplanetary Displacement

Archimedes' Principle: Any body immersed in a fluid is acted upon by an unbalanced upward force called the bouyant force, which is equal to the weight of the fluid displaced.

Recall that $\rho = \dfrac{m}{v}$ , where $\rho= density, m=mass,$ and $v=volume$ . We also know that $w=mg$ where $w=weight,m=mass$ and $g=gravitational~acceleration$ . So we have

$w=v \times \rho \times g$

Apply this to the fluid. Let $v_d$ be the displaced volume of fluid, $\rho _d$ be the density of the fluid displaced, $m_d$ be the mass of the fluid displaced, $w_d$ be the weight of the fluid displaced and $g$ be the acceleration due to gravity. We have

$w_d=bouyant~force=v_d \times \rho_d \times g$

Rearranging the equation, we get

$v_d=\dfrac{w_d}{\rho_d \times g}$

But we know that $w_d=m_d \times g$ , substitute

$v_d=\dfrac{m_d \times g}{\rho_d \times g}$

$v_d=\dfrac{m_d}{\rho_d}$

The gravitational acceleration has nothing to do with the volume displaced. It is because the volume displaced is equal to mass of the fluid displaced divided by density of the fluid displaced.

Note: It is assumed in the problem that the liquid is identical on both planets.

When an object floats in a liquid — as the wooden block does — the liquid displaced is equal in weight to the weight of the object. Since $\text{weight} = \text{mass}\times\text{gravity},$ it might be tempting to say that more liquid will be displaced in the higher gravity scenario (on Xylem). Mass, however, is independent of gravity, so regardless of which planet we're on, the same mass of liquid will be displaced!

It doesn't matter whether the gravity of "Xylem" has a greater pull force towards the center. Since both blocks are identical as well as the beaker of liquid, the same amount of liquid will be displaced no matter the gravity.

In the end, the liquid in Xylem would be attracted to Xylem at the same proportional rate as the wooden block, making gravitational attraction irrelevant

Intuitively; the floating effect depends on the difference between water and the wooden block density. Gravity doesn't affect their density, so it is the same on both planets.

Isn't it goes simple like this:- The volume of the fluid displaced is equal to the volume of the part of the body (say block here) submerged under the fluid.

The argument that "The same amount is displaced under the same scenario" is under the assumption that the densities (and thus the volumes) of the wooden block and water are not affected by the gravity change.

Since we are not explicitly given this assumption, truly the correct answer should be "Cannot be determined".

• Using Archimedes' Principle, we get (ρ B)(V T)g=(ρ F)(V F)g, Where ρ B is the density of the block, V T is the total volume of the block, g is any generic gravity, ρ F the density of the fluid, and V F is the volume of fluid being displaced by the block. With some algebraic manipulation, we get (V_F)= $\frac{(ρ_B)(V_T)}{(ρ_F)}$

Assuming the liquid is ideal and doesn't compress under any change in gravity, the volume displaced will remain the same.

The weight of the wooden block is more on the planet Xylem, but so is the weight of the liquid that has to be displaced, and by exactly the same proportion. Regards, David

Gravity acts constant over the entire unit, and as the wooden block is not thrown into water it will comply only force due to its weight. And in that case the boyant force comes into picture which obeys Archimedes principle that water displaced by the wooden block is equal to weight of block. Now if we equate the two cases acceleration due to gravity is cut out from the equation. And not does necessary work on water displacement.

Judging by the high standard of users, very simple logic would be the g won't affect the scenario. The densities of wood and water remain same, the downward wight mg increases per g increment but the buoyant force v(rho)g increases too, so it is still between m and (rho) the density.