Question #6ab42

2 Answers
Jul 22, 2015

Answer:

I found #5km# 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:
enter image source here

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#

Jul 22, 2015

Answer:

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 * 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")#