Question #6ab42

Consider the following:

The Law of Gravitation gives you:
`F_G=G(mM)/R^2` where `M=` mass of Earth.
If: radius Earth `r=6371km` then `r_1=r-10=6361km`

on the surface you have:
`g=9.81=GM/r^2`
In your case you have: `g_h=g_1`
`cancel(G)M/(r+h)^2=cancel(G)M_1/r_1^2` _ _ _ (1)
The problem is `M_1`!
If density of Earth is `delta=(mass)/(volume)` (assuming it constant through the entire planet);
`M/(4/3pir^3)=M_1/(4/3pir_1^3)`
`M_1=M(r_1/r)^3`

So substituting in (1)
`1/(r+h)^2=(r_1)/r^3` rearranging;
`(r+h)^2=r^3/r_1` taking the square root on both sides:
`r+h=6376`
`h=5km`

The idea of this problem is that you need to use the fact that the gravitational acceleration, `g`, decreases as you move away from the surface and as you move towards the center of the planet.

At the surface, the gravitational acceleration is given by this equation

`g = G * M/r^2`, where

`r` - the radius of the Earth;

At a distance equal to `h` above the surface, the gravitational acceleration will be

`g_h = G * M/(r + h)^2`

The trick here is to divide this value by the value of `g` at the surface

`g_h/g = (cancel(G) * cancel(M)/(r + h)^2)/(cancel(G) * cancel(M)/(r ^2)`

`g_h/g = r^2/(r + h)^2`

Now, when `h` is much smaller than `r`, you can write this expression like this

`g_h/g = [(r + h)^2/r^2]^(-1) = [((r + h))/r]^(-2) = (1 + h/r)^(-2)`

Use binomial expansion to write this as

`(1+h/r)^(-2) = 1 + (-2) * h/r + ((-2) * (-3))/(2!) * (h/r)^2 + ((-2) * (-3) * (-4))/(3!) * (h/r)^3 + ...`

`(1 + h/r)^(-2) = 1 - (2h)/r + 3(h/r)^2 - 4(h/r)^3 + ...`

Since `h` is much smaller than `r`, you can approximate this to be equal to

`(1 + h/r)^(-2) ~~ 1 - (2h)/r`

The gravitational acceleration at a height `h` above sea level will thus be

`g_h = g * [1-(2h)/r] " "color(blue)((1))`

At a distance `d` below the surface, the gravitational acceleration will be

`g_d = G * M_d/(r - d)^2`, where

`M_d` - the mass of the Earth that actually attracts the body.

SInce you're below the surface, not all the mass of the Earth will contribute to the value of `g_d`. This happens because your object is now attracted by a sphere that has a smaller radius, i.e. `(r-d)`.

Assuming that Earth's density is constant, you can write

`rho = M/V => M = rho * V`

`M_d = rho * underbrace(4/3pi * (r-d)^3)_(color(green)("volume of a sphere"))`

This means that `g_d` will be equal to

`g_d = G * (rho * 4/3pi (r-d)^cancel(3))/cancel((r-d)^2) = G * 4/3pi * rho * (r-d)`

Once again, divide this by the value of `g` at the surface to get

`g = G * M/r^2 = G * (rho * 4/3pi * r^(cancel(3)))/cancel(r^2) = G * 4/3pi * rho * r`

`g_d/g = (cancel(G * 4/3 pi * rho) * (r-d))/(cancel(G * 4/3pi * rho) * r) = (r-d)/r = 1 - d/r`

The gravitational acceleration at a distance `d` below the surface will thus be

`g_d = g * (1 - d/r) " "color(blue)((2))`

Now all you have to do is set these two equations equal to each other and solve for `h`

`color(blue)((1) = (2)) => g_h = g_d`

`cancel(g) * (1 - (2h)/r) = cancel(g) * (1 - d/r)`

This is equivalent to

`(2h)/cancel(r) = d/cancel(r) => h = d/2`

In your case, `d` is equal to 10 km, therefore

`h = "10 km"/2 = color(green)("5 km")`