Gravatational Potential Energy

In summary, Lemmy found a formula for the radius in terms of the mass and the gravitational constant. Can you use this information to solve for the mass of the satellite?
  • #1
Lemmy
22
0
Okay I understand somewhat of this topic of grav potential energy but i got stumped on this one question. I am to find the mass of a satellite orbiting Earth with the given variables of (F)(units: kN) which is the force the sattelite attracts the earth, and given the gravitational potential energy (-U). You are also givin the gravitational constant (G), and the Earth (m_e)

How could you find the mass (m) of the sattelite without a given radius in the equation U=-(G*m_e*m)/r ?

How would i rewrite this equation

i know if i was givin radius then i could write it as m=(U*r)/-(G*m_e)

any help would be great thanks
lemmy

here is the problem
"When in orbit, a communication satellite attracts the Earth with a force of F and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - U"
"Find the mass of the satellite."
"Take the gravitational constant to be G , the mass of the Earth to be m_e ."
 
Last edited:
Physics news on Phys.org
  • #2
Hmmm maybe F=(G*m*m_e)/r^2, i could possibly do r^2=(G*m*m_e)/F then r=sqrt((G*m*m_e)/F)
u think that's how to do radius?
 
  • #3
tell me if this sounds right, m=(U*sqrt((G*m*m_e)/F))/(G*m_e)
aww nm i don't know m :(
 
  • #4
Hi again Lemmy, please don't post multiple copies of the same question in different forums, thanks!

Regarding your work so far, you have found a formula for the radius in terms of the unknown mass and known quantities (m_e, G, F). Can you now find another formula that relates m and r so that you can plug in and solve for m?
 
  • #5
well U= -(G*m_e*m)/(r) and F=-(G*m_e*m)/r^2 is all that comes to mind right now
 
  • #6
Lemmy, you keep going in circles. I can't just tell you the answer, but it is staring you right in face. Think about what equations you have used already and what equations you haven't used yet. Hint: you know r in terms of m, so look back at your first post.
 
  • #7
what do you mean i know r in terms of m? and i still don't understand where for would apply,tricky part is I'm not allowed to have r in my answer
 
Last edited:
  • #8
i'm trying to find the equation to solve this problem
 
  • #9
Lemmy, in post number 2 you give a formula for radius r in terms of mass m, G, m_e, and F. This formula came from the force equation. In post one you give a formula for mass m in terms of radius r, G, m_e, and U. This formula came from the potential energy equation. Can you somehow combine the two?

Edit: Look at post 3, you've already done it!
 
  • #10
m=(U*sqrt((G*m*m_e)/F))/(G*m_e)
thats not it because i can't use m in the (U*sqrt((G*m*m_e)/F)) because m is not given in both equations they both involve m and r, and both m and r are not given in both equations, so i can't use what i posted in the third post
 
  • #11
hehe i tried submitting the third post answer and its wrong :(
 
  • #12
Lemmy, you have one equation there involving only m, you can solve it for m in terms of just G, m_e, F, and U.

Let me write it for you:
[tex]
m = \frac{U \sqrt{G m_e m}}{\sqrt{F}}\frac{1}{G m_e},
[/tex]
can you see how to find m?
 
Last edited:
  • #13
shouldn't the (U*sqrt((G*m_e*m)/F))) all be in the numerator?
 
  • #14
Yes, sorry, I fixed it. Can you solve that for m? Hint: get all the m terms on one side.
 
  • #16
thing is physics monkey i can't change the left side term, because the problem has everything set to m= already and i can't change left, only answer i can submit is the right term
 
  • #17
Yes you can, Lemmy. That's how solving equations works. Maybe what you end up with is [tex] \sqrt{m} [/tex] equals something you know, so then you just take the square of both sides to find what m equals.
 
  • #18
oh hmm i see, okay i'll try it, my algebra skills suck though
 
  • #19
okay i solved and got m=(U^2)/(F*G*m_e)
can you check my algebra? i hope this is right *crosses fingers*
 
  • #20
Looks fine to me.
 
  • #21
okay going to give this a whirl thank you soo much for your help physics monkey, greatly appreciated
 
  • #22
woooo got it right thanks physics monkey
 

Related to Gravatational Potential Energy

What is gravitational potential energy?

Gravitational potential energy is the energy that an object possesses due to its position in a gravitational field. It is the potential for an object to do work as a result of its position in a gravitational field.

How is gravitational potential energy calculated?

Gravitational potential energy is calculated by multiplying the mass of an object by the acceleration due to gravity (9.8 m/s^2) and the height of the object above a reference point. The formula is PE = mgh, where PE is potential energy, m is mass, g is acceleration due to gravity, and h is height.

What factors affect an object's gravitational potential energy?

The two main factors that affect an object's gravitational potential energy are its mass and its height above a reference point. The greater the mass and the higher the object is above a reference point, the greater its gravitational potential energy will be.

Is gravitational potential energy conserved?

Yes, gravitational potential energy is a type of potential energy, which means it is stored energy that can be converted into different forms. It follows the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed.

How is gravitational potential energy related to kinetic energy?

Gravitational potential energy and kinetic energy are two forms of energy that are related to each other. As an object falls, its potential energy decreases while its kinetic energy increases. At the point of impact, all of the potential energy is converted into kinetic energy. The relationship between the two can be described by the equation PE + KE = constant.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
991
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
914
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
14
Views
1K
Replies
9
Views
384
Back
Top