Gravitational motion usingly earths radius

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Homework Help Overview

The discussion revolves around determining the velocity of a communication satellite in circular orbit around the Earth at a height of 36,000 km. The problem is framed within the context of gravitational motion and involves Newton's law of gravitation, with specific constraints on the numerical values that can be used in the calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between gravitational force and circular motion, attempting to set equations equal to find the satellite's velocity. There are discussions about the appropriate values to use for mass and radius, with some questioning the use of the satellite's height directly in calculations.

Discussion Status

Some participants have offered guidance on how to approach the problem using gravitational equations, while others express confusion about the constraints of the problem. There is an ongoing exploration of whether certain values can be used and how to interpret the problem's requirements.

Contextual Notes

Participants note that the problem explicitly states that only the radius of the Earth and the acceleration due to gravity at the surface can be used, leading to confusion about the relevance of the satellite's orbital height. There is also mention of the potential need for additional information, such as the orbital period of the satellite.

zelphics
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Homework Statement


A communication satellite is in circular orbit around the Earth at 36,000 km above the earth. Starting with Newton's law of gravitation, determine the velocity of the satellite. The only numerical values you are allowed to substitute in solving this problem are the radius of the Earth and the acceleration due to gravity at the surface of the earth

Homework Equations


(mv^2)/(r)

F = ((Gm1m2)/(r^2))


The Attempt at a Solution



I set them equal and got the equation

v^2 = (Gm2)/(r)

i don't know what to do, from here.
 
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That's correct. Now plug in the right values for

m2 and r into:

v^2 = (Gm2)/(r)

and solve for v.

m2 is the mass of the earth. look it up online or in your text.

r is the distance from the center of the earth.

ie r = radius of the Earth + 36,000*10^3

Look up the radius of the earth.

EDIT:

Sorry zelphics... I didn't read the whole post.

you can use g = Gm2/r^2

so m2 = g*r^2/G where r is the radius of the earth...

then plug this in for m2 in:

v^2 = (Gm2)/(r)
 
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You are allowed to use a numberical value for the acceleration due to gravity at the surface of the Earth, which is of course one Earth radius from the center of the Earth. What does Newton's law of gravitation (along with Newton's second law of motion) tell you the acceleration due to gravity is at a distance of one Earth radius from the center of the Earth? What does that mean in terms of the satellite's motion?
 
learningphysics said:
m2 is the mass of the earth. look it up online or in your text.

He can't use the mass of the Earth or the value for G in this problem. He has to use the given satellite orbital radius, the radius of the Earth, the acceleration due to gravity at the surface of the Earth, and nothing else.
 
Learningphysics: I can't just plug in the mass of the earth...

D H: Newtons second law is F = ma, I have no idea - I am really confused.
 
D H said:
He can't use the mass of the Earth or the value for G in this problem. He has to use the given satellite orbital radius, the radius of the Earth, the acceleration due to gravity at the surface of the Earth, and nothing else.

Yes, sorry about that. :redface:
 
zelphics said:
Learningphysics: I can't just plug in the mass of the earth...

D H: Newtons second law is F = ma, I have no idea - I am really confused.

you can use g = Gm2/r_earth^2

where r_earth is the radius of the earth... solve for m2 here... plug it into your equation.
 
zelphics said:
Learningphysics: I can't just plug in the mass of the earth...

D H: Newtons second law is F = ma, I have no idea - I am really confused.

The law of gravitation describes the force due to gravity. The force due to gravity translates into the acceleration due to gravity after dividing the force by the mass, which you can do by virtue of Newton's second law.
 
learningphysics said:
you can use g = Gm2/r_earth^2

where r_earth is the radius of the earth... solve for m2 here... plug it into your equation.

Solve for the product Gm2. He can't use G.
 
  • #10
Edit
 
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  • #11
D H said:
Solve for the product Gm2. He can't use G.

Thanks D H. I'm not thinking straight today.
 
  • #12
aaron?

from my physics class at FM?

NO WAY...it's zach!
 
  • #13
hahahaha!

I knew it was someone from my class

it was the same exact problem.

Gm2= g at Re Re^2

thats what you are looking for

Me and Kagan just figured that out we hope its right.
 
  • #14
do you guys have aol or something?

i am soo confused - I am not even going to sleep tonight =[
 
  • #15
Yeah we were just on AOL trying to do them all.

We think we got all of them except the very last part(c) of #6.

We gave up and got off but I'm on just re-writing all of them over on nice paper.
 
  • #16
im still on #3, are there any pages in the book that are helpful?

i read it a million times...
 
  • #17
Umm not that I could find...

I re-read it tons too.

But I think I can do most of these mayybe.

I can try to help you but I won't guarantee they are all perfect.

On #3 you are really close you have v^2=Gm2/r

GMe=g at Re Re^2 plug that in

That me and Kagan found in our notes from Oct. 22
 
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  • #18
that would be really helpful if you could help me, but i don't want to keep you up - maybe we shouldn't use this forum as a chat room though =p
 
  • #19
Aaron, please do not give out answers to homework problems. Doing so doesn't help the person who asked the question learn anything.

Even more important, please do not give out wrong answers. Doing so hinders learning.
 
  • #20
Well maybe I did do it wrong but i had to find a way to do it without the 36,000 since we aren't even allowed to use that.
 
  • #21
Uh oops sorry then I didn't know.

themastachief591 add me if you want

I edited my post no more answer and no more wrong answer there really sorry.
 
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  • #22
its not your fault, its a tricky question...and you got farther than me - so it was helpful!
 
  • #23
AaronPrahst said:
Well maybe I did do it wrong but i had to find a way to do it without the 36,000 since we aren't even allowed to use that.

I think you have to use 36,000*10^3m
 
  • #24
i added you, my aol name is the same as it is on here.
 
  • #25
Again I apologize for doing it wrong but the question does say that the only numerical values we can use are the radius of the Earth and the acceleration due to gravity at the Earth's surface.

So wouldn't that mean we can't use 36,000x10^3m?
 
  • #26
You must have been given some information about the satellite -- a 24 hour orbit perchance?

BTW, thanks for deleting that post. OTOH, now I can't help you find your error. You somehow calculated the velocity of a satellite orbiting zero meters above the surface of the Earth.
 
  • #27
AaronPrahst said:
Again I apologize for doing it wrong but the question does say that the only numerical values we can use are the radius of the Earth and the acceleration due to gravity at the Earth's surface.

So wouldn't that mean we can't use 36,000x10^3m?

I think they just meant the constants... seems strange they would give the 36,000 in the question if you couldn't use it. The speed depends on the distance from the Earth's center... so we'd have to use that somehow...
 
  • #28
Yeah I did mess up big time but still our physics prof. gave us this question and so maybe the question is flawed.

I mean it pretty clearly says we can't use 36,000km.

And yeah how do we get the speed without using 36,000km? Its impossible.

When we get this wrong I'm going to go complain to him.
 
  • #29
The problem is indeed soluble knowing the orbital period, the radius of the Earth, and the acceleration due to gravity at the surface of the Earth. What is the radial acceleration of an object in uniform circular motion at radius r and angular rate \omega?
 
  • #30
The question said it was a communication satellite. Did your instructor tell you that communication satellites typically have an orbital period of 24 hours?
 

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