Question

1) How high must you be above the surface of the earth so that the gravitational force is half of that near the surface.

2)What is the height of the geostationary satellite?

Answer 1

If the gravitational force on the surface of the ground is \(F_1\), and the distance to and the radius of the earth is \(r\)
\[F_1 = k \frac{1}{r^2} ------- (1)\]

(note that \(k = GmM\))

if at \(R\) distance the force is halved,
\[\frac{F_1}{2} = k \frac{1}{R^2}------(2)\]

you know the value of \(r\) so from (1) and (2) get an expression for R with use of r

for the second one,

a geostationary sattelite will have the same angular velocity of the earth. you know the angular velocity of the earth right ;) ?..

use following equations to find the radius of the sattelite to have that amount of an angular velocity

\[F= \frac{GMm}{r^2}\]\[F = mr\omega^2\]

Answer 2

hmm. So R= .5^2 ?

Answer 3

wait do I solve for R instead in equation 2?

Answer 4

I was thinking solving for solving for F1, then plug it in equation (2) which all I need to do is divide my equation (1) by 2

Answer 5

yes finding F_1 is not neccessary

Answer 6

I need help. I got r to be 6.371*10^6

Answer 7

6.371m*10^6

Answer 8

do I set equation (1) and (2) equal?

Answer 9

I really need help

Answer 10

you can remove k and F1 from both equation by (1)/(2) so that only r and R will be left.. since youve found r you'll find R

Answer 11

I get R=r/sqrt(2)

Answer 12

On the seconds question I set them equal as well right.. I am a bit confuse on second question because finding the satellite near the surface of the earth, you set F_g = F_c where F_c is the m times a_c (centripetal acceleration). Why do we use momentum for this problem?

Answer 13

where did we use momentum here o_O?

Answer 14

I did not meant to say that. It was really late.