# Question 6ab42

Jul 22, 2015

I found $5 k m$ BUT: I had to use a given value for the radius of the Earth (not given) and I supposed the density of Earth being uniform throughout the entire planet. Probably my method is not completely reliable! Check it!

#### Explanation:

Consider the following:

The Law of Gravitation gives you:
${F}_{G} = G \frac{m M}{R} ^ 2$ where $M =$ mass of Earth.
If: radius Earth $r = 6371 k m$ then ${r}_{1} = r - 10 = 6361 k m$

on the surface you have:
$g = 9.81 = G \frac{M}{r} ^ 2$
In your case you have: ${g}_{h} = {g}_{1}$
$\cancel{G} \frac{M}{r + h} ^ 2 = \cancel{G} {M}_{1} / {r}_{1}^{2}$ _ _ _(1)
The problem is ${M}_{1}$!
If density of Earth is $\delta = \frac{m a s s}{v o l u m e}$ (assuming it constant through the entire planet);
$\frac{M}{\frac{4}{3} \pi {r}^{3}} = {M}_{1} / \left(\frac{4}{3} \pi {r}_{1}^{3}\right)$
${M}_{1} = M {\left({r}_{1} / r\right)}^{3}$

So substituting in (1)
$\frac{1}{r + h} ^ 2 = \frac{{r}_{1}}{r} ^ 3$ rearranging;
${\left(r + h\right)}^{2} = {r}^{3} / {r}_{1}$ taking the square root on both sides:
$r + h = 6376$
$h = 5 k m$

Jul 22, 2015

Here's how you can solve this one without knowing Earth's radius.

#### Explanation:

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

$g = G \cdot \frac{M}{r} ^ 2$, where

$r$ - the radius of the Earth;

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

${g}_{h} = G \cdot \frac{M}{r + h} ^ 2$

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)

${g}_{h} / g = {r}^{2} / {\left(r + h\right)}^{2}$

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

${g}_{h} / g = {\left[{\left(r + h\right)}^{2} / {r}^{2}\right]}^{- 1} = {\left[\frac{\left(r + h\right)}{r}\right]}^{- 2} = {\left(1 + \frac{h}{r}\right)}^{- 2}$

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 + ...

${\left(1 + \frac{h}{r}\right)}^{- 2} = 1 - \frac{2 h}{r} + 3 {\left(\frac{h}{r}\right)}^{2} - 4 {\left(\frac{h}{r}\right)}^{3} + \ldots$

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

${\left(1 + \frac{h}{r}\right)}^{- 2} \approx 1 - \frac{2 h}{r}$

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

${g}_{h} = g \cdot \left[1 - \frac{2 h}{r}\right] \text{ } \textcolor{b l u e}{\left(1\right)}$

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

${g}_{d} = G \cdot {M}_{d} / {\left(r - d\right)}^{2}$, 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. $\left(r - d\right)$.

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

$\rho = \frac{M}{V} \implies M = \rho \cdot V$

${M}_{d} = \rho \cdot {\underbrace{\frac{4}{3} \pi \cdot {\left(r - d\right)}^{3}}}_{\textcolor{g r e e n}{\text{volume of a sphere}}}$

This means that ${g}_{d}$ will be equal to

${g}_{d} = G \cdot \frac{\rho \cdot \frac{4}{3} \pi {\left(r - d\right)}^{\cancel{3}}}{\cancel{{\left(r - d\right)}^{2}}} = G \cdot \frac{4}{3} \pi \cdot \rho \cdot \left(r - d\right)$

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

$g = G \cdot \frac{M}{r} ^ 2 = G \cdot \frac{\rho \cdot \frac{4}{3} \pi \cdot {r}^{\cancel{3}}}{\cancel{{r}^{2}}} = G \cdot \frac{4}{3} \pi \cdot \rho \cdot r$

${g}_{d} / g = \frac{\cancel{G \cdot \frac{4}{3} \pi \cdot \rho} \cdot \left(r - d\right)}{\cancel{G \cdot \frac{4}{3} \pi \cdot \rho} \cdot r} = \frac{r - d}{r} = 1 - \frac{d}{r}$

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

${g}_{d} = g \cdot \left(1 - \frac{d}{r}\right) \text{ } \textcolor{b l u e}{\left(2\right)}$

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

$\textcolor{b l u e}{\left(1\right) = \left(2\right)} \implies {g}_{h} = {g}_{d}$

$\cancel{g} \cdot \left(1 - \frac{2 h}{r}\right) = \cancel{g} \cdot \left(1 - \frac{d}{r}\right)$

This is equivalent to

$\frac{2 h}{\cancel{r}} = \frac{d}{\cancel{r}} \implies h = \frac{d}{2}$

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

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