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Satellite moving in a stable circular orbit 
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#1
Dec1509, 03:12 PM

P: 6

1. The problem statement, all variables and given/known data
A 600kg satellite moving in a stable circular orbit about the Earth at a height of 4000km (G=6.67x10^11 NM^2/kg^2, Re=6380km, Me=5.98x10^24kg). Calculate the speed of the satellite at that height. Calculate the orbital period (T), the time for one revolution Calculate the gravity at height h=4000 km. 2. Relevant equations Gravitational Force: F=Gm1m2/r^2 centripetal acceleration: a= v^2/r Period (circular motion): T=2(pie)r/v 3. The attempt at a solution calculated a gravitational force of 2221179013, plugged it into F=ma, which I came out as 2221179013=600a. Got an acceleration of 3701965.021=v^2/r, which came out to v2= 3701965.02*10380. After taking the square root of that, I finished with v=196026.52m/s, which is off from the correct answer which is 6198.9 m/s. Where did I go wrong, how can I take the right steps? Thanks so much. 


#2
Dec1509, 05:01 PM

P: 6

I've been working on this question for a couple hours now and I can't seem to get a different answer. Is it possible that the answer given in the study guide is incorrect?



#3
Dec1509, 07:22 PM

P: 6

I just need help for the velocity part. I have a final on this tomorrow, can someone please help me find the error in my calculation? Thanks.



#4
Dec1509, 07:29 PM

HW Helper
P: 2,155

Satellite moving in a stable circular orbit
Your calculation is pretty much meaningless to me because you don't have any units attached to your numbers. I can't tell what error you might have made unless you have the proper units on each value.
For what it's worth, I was easily able to confirm the answer in the study guide, about 6.2 km/s. 


#5
Dec1509, 08:00 PM

Mentor
P: 15,065

You nailed it diazona (well, almost. It would have helped if ronartest had shown some work.)
Units, ronartest! These quantities are not just numbers. 


#6
Dec1509, 09:38 PM

HW Helper
P: 2,155

Yay! Finally someone who cares as much as I do about units ;)



#7
Dec1509, 09:45 PM

P: 6

Haha, yeah I'm really sorry about that. To be honest, I'm not really sure of the units on anything except the initial numbers. That said for the gravitational force I multiplied NM^2/kg^2 by kg by kg divided by km, whatever that comes out to. I guess the acceleration would be m/s^2. Is it not possible to understand the steps I am taking? Would it be possible to explain the steps that you took to achieve your answer? Again, I'm sorry for not being more clear, and I greatly appreciate your interest.



#8
Dec1509, 10:54 PM

HW Helper
P: 2,155

Would you attempt to do this problem without knowing how to multiply? Because that's just what you did. You really can't expect to be able to solve physics problems without knowing how to do basic math, and that includes both numbers and units.
So what are the units on the gravitational force you calculated? [tex]F = G\frac{m_1 m_2}{r^2}[/tex] You're right that G has units of N m^2/kg^2 (it's lowercase m for meters, not capital M), and both m_{1} and m_{2} have units of kilograms, and that r has units of kilometers (but note that r is squared in the denominator). How would you figure out the result of multiplying (N m^2/kg^2) * kg * kg / km^2? 


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