# Homework Help: GRE 75, orbits, satellites and their periods

1. Sep 20, 2004

### quantumworld

Here is a question that I am having a hard time understanding.

75) the period of a hypothetical earth satellite orbiting at sea level would be 80 minutes. In terms of the earth's radius Re, the radius of a synchronous satellite orbit ( period 24 hours) is mostly:
(A) 3 Re
(B) 7 Re
(C) 18 Re
(D) 320 Re
(E) 5800 Re
the answer is (B) 7 Re, but how is that, and I was surprised that it takes the earth 24 hours to go around itself, but a satellite could make it in only 80 minutes :surprised . But most importantly is how did they get 7 Re for the answer, and I actually don't understand the question itself...

Thank u so much!

2. Sep 20, 2004

### K.J.Healey

For the 80 minutes (or 1.333hours) just think how fast something would have to be traveling around the earth at sea level (ignore terrain) to stay in orbit (avoid crashing to the ground). It has to be crazy fast.

The ultimate equation to remember for that is that
R1^3 / T1^2 = R2^3 / T2^2

So r^3/T^2 is equal to some constant
therefor r^3 is proportional to T^2
So for the first part
Re^3 = k(constant)*(4/3 hrs)^2
solve for k
k = 9/16 Re^3

so plug that in for the higher orbit

X^3 = (9/16 Re^3) * (24 hours)^2
X^3 = (24*24*9/16)Re^3 = (1.5*24*9)Re^3 = 1.5*216*Re^3 = 324 Re^3
this is the tricky part I would guess. Its not exactly 7
you need to take the cubed root of 324. I would, here , just cube each of the answers and see wthich is the closest, which is 7 at (343).

Man am I DREADING this test in november.

3. Sep 21, 2004

### nrqed

Healey01 already provided the solution. It's important to know kepler's 3rd law: P^2 = a^3.

${ 4/3 h \over 24 h}^2 = {R_E \over N R_E}^3$
$18^2 = N^3$ or $N= 18^{2/3}$. It's then easy to get the answer (for example, by writing $18^2 = 27 \times 8 \times 1.5$ so the cube root is a bit more than 6).