Gravitational motion usingly earths radius

In summary, a communication satellite is in circular orbit around the Earth at 36,000 km above the earth. The student is asked to determine the velocity of the satellite using Newton's law of gravitation and only the radius of the Earth and the acceleration due to gravity at the Earth's surface as numerical values. After some discussion and attempts at solving the problem, the student is still unsure of how to proceed and is seeking further guidance.
  • #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.
 
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  • #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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