How Far Above Earth is the Satellite?

Rbethell16
Messages
10
Reaction score
0
1. Homework Statement
The radius of the Earth is 6.86 e6 m and the mass of the Earth is 5.98 e24kg. There is a satellite floating in space at a point where the Earth's gravitational acceleration is 3.2 m/s^2.

How far above the Earths surface is the Satellite?
2. Homework Equations
GMm/R^2 = F


3. The Attempt at a Solution
So I did 3.2 = GM/R^2...R = 1164480.51. But that isn't right - could someone please guide me as to how I should go about setting up the equation
 
Physics news on Phys.org
That is the correct equation. Why do you think your answer is wrong?
 
I didn't check the calculation, but let's assume you did it correctly and R is indeed 1164480.

Then whát distance exactly is 1164480 (and in what units? Always state your units!).
Is that the distance you were asked to find?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top