Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radi

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 FR 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 FR/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.

Explanation / Answer

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

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