Question

Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radial tunnel through its center. Also assume we can position an apple anywhere along the tunnel or outside the sphere. Let F_R be the magnitude of the gravitational force on the apple when it is located at the planets surface. How far from the surface is there a point where the magnitude of the gravitational force on the apple is F_R/2 if we move the apple (a) away from the planet and (b) into the tunnel? Express you answer in terms of the variables given.

Answer 1

a)

Let m be the mass of the apple and M be the mass of the Earth

Radius of the planet is is R

a ) Gravitational force is F_R = G Mm / R_e^2

F_e = GMm / (R_e +h)^2

Given that F_e = 0.7 F_R

= 1/ v2 F_R

GMm / (R_e +h)^2 = 1/ v2 * G Mm / R_e^2

v2 R_e ^2 = ((R_e +h)^2)

(0.7 ) R_e^2 = (R_e +h)^2

0.83 R_e = (R_e +h)

h = 0.163 R_e

b)

  its because its a sphere. and in this case the mass affects the gravity. just think about it. if you got a two spheres, one with 1 meter of radius and one with 2 meters of radius their volume would be this

first: 4/3*3,14*1^3= 4,18m^3

second 4/3*3,14*2^3=33,49m^3

the diameter of the second sphere is only the double of the first one but its volume (therefore its mass and its gravity) is more than 8 times bigger.

I'm too sleepy to calculate it for you but i think you should recognized your faulty by now

edit: okay i write down how to calculate it, because i'm that awesome :)

the volume of the sphere is this:

4/3*Pi*r^3

in your case you can think of this planet radius as 1.

so your answer is

4/3*Pi*(1-x)^3=(4/3*Pi)*0,6
(1-x)^3=0,6

1-x=0,84

the x is the distance what the apple falls in this tunnel so your answer would be r=0,84R