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Plant X has a radius of 5000 km and is composed of two layers. The first inner layer ranges from the centre to 2000 km from centre, it's density is 8 kg / dm^3. The second layer ranges from 2000 km to 5000 km from the centre, it's density is 4 kg/ dm^3. What is the gravitational acceleration inside the planet at 4000 km from it's centre? |
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The key point to solve this problem is you should know the material on a spherical shell has no gravitational effect inside the shell. So to calcite the gravitational acceleration, you just need to consider the contribution from mass inside the $4000km$ sphere. I think you should already know how to calculate the gravitational acceleration of a planet has constant density: $Gravitation Force = Gm\frac{\iiint\rho(r‘)dV}{r^2}$ where $r$ is the distance of object $m$ from the center of planet, and $\rho(r')$ is the density of planet, it's a function of $r'$ (as in your problem). We know the gravitation force on the object $m$ is $Force = ma$, so the acceleration is $a = G\frac{\iiint\rho(r‘)dV}{r^2} = G\frac{4\pi\int\rho(r')r'^2dr'}{r^2}$ In your case, $r = 4000km$, using the conclusion stated in my first paragraph, and the density stated in your problem, we should have this: $\iiint \rho(r')dV = 4\pi\int\rho(r')r'^2dr' = 4\pi(\int_0^{2000km}\rho_1r'^2dr' + \int_{2000km}^{4000km}\rho_2r'^2dr')$ where $\rho_1 = 8kg/L$ and $\rho_2 = 4kg/L$ as stated in your problem. Also, there is a simple way to do this problem. Since the density here $\rho_1$ is twice of $\rho_2$, so you can think like this: the gravitational force of this single planet can be effectively equal to a total effect of two planets have same densities, but one planet only has $2000km$ radius, and the other has $4000km$ radius, so calculate the acceleration of both planets at the radius of $4000km$, and then combine these two to get the total acceleration. |
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Do you know how to do the problem for a planet of constant density? You really need to do this problem before you try the layered planet. EDIT: Matt has answered that he knows how to "calculate" g for the constant density case, but this answer troubles me because I wasn't thinking of a calculation when I asked if he knew how to "do" the problem. I'm going to tell him how I "do" the problem and hope it will help him to "calculate" the answer. For the planet Earth, the gravitational force inside the planet (that is, if you drilled a tunnel to the center of the earth) is linear with depth. In other words, when you get halfway to the center, g is 4.9 m/sec^2. I know this for all kinds of reasons but you should be able to convince yourself it is true before you go any further. When you are above the surface, the force is 1/r^2. So for example when you are 4000 miles above the surface, g is 2.45 m/sec^2. These are the two cases you need to know, and as Qmechanic has correctly suggested, the problem as given should be solved as the superposition of these two cases. |
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