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.

  • #31
He didn't tell us that.
 
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  • #32
D H said:
The question said it was a communication satellite. Did your instructor tell you that communication satellites typically have an orbital period of 24 hours?

Would they be allowed to use that in the problem? still a numerical value...
 
  • #33
Your instructor must have told you something. Please write the problem exactly as it was given to you.
 
  • #34
D H said:
Your instructor must have told you something. Please write the problem exactly as it was given to you.

The 36,000km.
 
  • #35
Learningphysics, you have a PM from me. I need to log out for now.
 
  • #36
He stated the problem perfectly.

I just had a thought...we can't substitute 36,000km, but if he already gave it to us like that then maybe that means he just substituted it for us, so maybe we are supposed to have it already in all our equations for h.
 
  • #37
If you guys are allowed to use the 36,000km...

m1g = Gm1m2/r_earth^2 (m1 = arbitrary mass. m2 = mass of earth)

so

g = Gm2/r_earth^2

where m2 is the mass of the earth...

So Gm2 = g*r_earth^2

Then plug this into the equation (as solved in the original post in this thread)

v^2 = Gm2/r

so

v^2 = g*r_earth^2/r

v^2 = g*r_earth^2/(r_earth + 36000*10^3)

plug in g=9.8m/s^2. r_earth = 6370*10^3m

at least this gives you the numerical answer you need according to the initial problem descripton... so no matter what method you use (ie a way to solve without using the 36000*10^3m), this is the numerical result you should get...
 
Last edited:
  • #38
Well thanks for trying to help us but I guess our teacher is just bad and we are screwed for now.

One question though I thought Gm1=(g at Re)timesRe^2

You just have down Gm1=(g at Re)^2
 
  • #39
AaronPrahst said:
Well thanks for trying to help us but I guess our teacher is just bad and we are screwed for now.

One question though I thought Gm1=(g at Re)timesRe^2

You just have down Gm1=(g at Re)^2

I wrote

g = Gm2/r_earth^2

Gm2 = g*r_earth^2

The "*" is multiplied... so Gm2 is g multiplied by Re^2... just as you wrote...
 
  • #40
Oh ok I understand that now sorry I just mis-read that.

Thanks again.
 
  • #41
AaronPrahst said:
Oh ok I understand that now sorry I just mis-read that.

Thanks again.

Did you try the formula? What number do you get for velocity?
 
  • #42
Hmm well I plugged it in and got 1.32 m/s. Uh but I probably did something wrong...
 
  • #43
AaronPrahst said:
Hmm well I plugged it in and got 1.32 m/s. Uh but I probably did something wrong...

v^2 = g*r_earth^2/(r_earth + 36000*10^3)

v^2 = 9.8*(6370*10^3)^2/(6370*10^3+36000*10^3)

I get v = 3063.5m/s
 
  • #44
Ah ok yeah I got that now...I think I accidentally didn't square the Re on top.
 

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