Finding the mass of a planet knowing the density

Question

Need help:

Problem:

The planet solaris has a spherical form and a fixed core with radius R. The whole planet is covered in an ocean with the depth $D$. The floating substance that the ocean is made of, has the denisty

$$\frac{\alpha}{R+h}$$

in height $h$ over the ocean bottom, for $0\leq h\leq D$, where $\alpha$ is a constant

Find the mass of ocean that covers the planet

attempt:

I know that mass is volume times density, and since we are working with a sphere with radius R this will yield us the following

$$V=\frac{4\pi R^3}{3}$$ since $R=r$

which should give us the following integral

$$M=\int^D_0\frac{\alpha}{R+h}dV$$

where $dV=4\pi R^2dR$

$$M=\int^D_0\frac{\alpha}{R+h}\cdot4\pi R^2dR=4\pi\alpha\int^D_0\frac{R}{1+h}dR$$

$$4\pi\alpha\bigg[\frac{R^2}{2+2h}\bigg]^D_0=4\pi\alpha\bigg(\frac{D^2}{2+2h}-0\bigg)$$

$$\therefore{M=\frac{2\pi\alpha D^2}{h+1}}$$

is this correct?

Answer 1

You are integrating a bunch of spherical shells. At a radius $r$ the shell has area $4\pi r^2$ and thickness $dh$, for a volume $4\pi r^2 dh$. We have $r=R+h$ so the volume is $4\pi (R+h)^2 dh$ and the mass is $4\pi (R+h)^2 \frac \alpha{R+h}dh$ When you get cancellation you must be doing something right. We therefore want $$\int_0^D4\pi (R+h)\alpha \;\text dh$$

Answer 2

Hint: Firstly, using $R$ as the variable of integration is incorrect since it conflicts with the constant $R$. Secondly, your calculation of $dV$ is not quite correct; what you want is for it to be in terms of $h$ and $dh$. You are correct to think of it as the surface area of a sphere multiplied by a differential in radius.

Recall that you chose the limits of integration to be $0$ and $D$ because those are the bounds of the variable $h$. This means you desire a $dh$ at the end of your integrand. Think about what $h$ really means in terms of the geometry of the sphere.

Hope this helps.