Satellite orbiting Mars

Hi, I'm having trouble with this question:

Mars rotates on its axis once every 24.8 hours.What is the speed of a geosynchronous satellite orbiting Mars?

I thought I could use the equation:

v= the square root of Gm/r

however I've realized that the hight of the satellie's orbit has not been given so I cannot use this eqation.

If anyone can help me with this that would be great

Thanks!!
 

Doc Al

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You'll need to figure out the height of the satellite's orbit (or at least its distance from the center of the planet).

What's the angular speed? What distance from the planet's center will allow a satellite to stay in orbit with such an angular speed?

Answer these first, then you'll be able to determine the speed of the satellite.
 
so i've used the equation 2(3.14)/T (sorry I don't have anything for the symbols for pi) and I put in:

2(3.14)/(24.8h * 3600s) and I got the angular velocity to be 7.09x10^-5 rads per second.

now I'm unsure how to use this to find height of the orbit
 

Doc Al

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Use what you know about orbits. Apply Newton's 2nd law where gravity is the force. (This is related to the equation you quoted in your first post.) How does linear speed relate to angular speed?
 
Oh thank you so much, using the angular velocity i was able to find the radius of the orbit to be 20423153.3 m and then i was able to use the original equation that I posted in my 1st post to find the speed to be 1.44km/s. Thanks so MUCH for your help!!
 
and i wanna ask where i can get the mass of Mars. and after we got the v, how we can get the altitude of a geosynchronous satellite orbiting Mars. i use the equation r+h=2pi/(vT). but it is wrong.
 

Doc Al

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The mass and radius of Mars can be looked up. (Google it!)
 
and how about the altitude of a geosynchronous satellite orbiting Mars. how can i get it. which equation shall i use
 

Doc Al

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and how about the altitude of a geosynchronous satellite orbiting Mars. how can i get it. which equation shall i use
You can solve for the radius of the orbit using a combination of Newton's 2nd law, the law of gravity, and your knowledge of centripetal acceleration.
 
i used it as the sunny said above,20423153.3 m. i type this answer in my online quiz. it said wrong. Once again,i use the v which sunny get above, 1.44km/s, to get the altitude which is 20471847.13. when i type this answer. it is still wrong. once again. i use 20471847.13-20423153.3=48693.83376. but it is still wrong. i dont know why.
 

Doc Al

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i used it as the sunny said above,20423153.3 m.
That's the radius of the orbit, not the altitude. What's the radius of mars?
 
the radius of mars is 3397000m but i tried adding that to the 20423153.3m to get 23820053, however that is wrong as well.
 
ohhh i see what i was doing wrong, i was sippose to subtract the mass of mars, thanks!!!
 
Okay, i have a similar question to Sunny's. However i have to find the distance away in kilometers from the surface of Mars that a geosynchronous satellite must be placed in order to stay in the same place above the planet.

What should i do? I'm confused on which equation would be the best to use.

So i have that Mars Rotates on it's axis every 24.8 hours.

Could I use the equation V= the square root of (G x mass of Mars) over the radius?

Then i could make the velocity 0, and solve for the radius?
 
And what is angular speed? how do i find what angle the satellite is at?
 
hmm or should i do (24.8^2)/Radius^3=4(3.14)^2/GM
 
Ugh i'm completely lost. Newtons 2nd law would only be helpful if i had the mass of the satellite right?
 
Last edited:
Ah i made the equation R=((G x M x T^2)/(4pi^2))^(1/3)

That gets me the radius from the center of Mars to the Satellite, which i then subtracted the radius of mars alone to find the distance from the surface. Does that make sense? I got a similar answer to the kids before me.

-OH physics....
 

Doc Al

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Ah i made the equation R=((G x M x T^2)/(4pi^2))^(1/3)

That gets me the radius from the center of Mars to the Satellite, which i then subtracted the radius of mars alone to find the distance from the surface. Does that make sense? I got a similar answer to the kids before me.
That sounds good to me. (Make sure you know how to derive this equation from basic principles: Newton's 2nd law and centripetal acceleration.)
 

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