Space shuttle releases a sattelite into a circular orbit

AI Thread Summary
The discussion focuses on calculating the speed at which a space shuttle must move to release a satellite into a circular orbit 680 km above Earth. The participants explore the gravitational force equation and the relationship between the shuttle's velocity and the satellite's orbital velocity. Initial calculations using the formula yield an incorrect speed of 237690 m/s, while the correct answer is 7520 m/s. A key realization is the importance of using consistent units, specifically converting kilometers to meters. The conversation emphasizes the need for careful unit management in physics problems to avoid errors.
jmcmillian
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The space shuttle releases a sattelite into a circular orbit 680 km above the Earth. How fast must the shuttle be moving (relative to the Earth's center) when the release occurs?


I know that G (m*m(earth)/r^2) = m (v^2/r)...with r = r(earth) + h. In this equation, r=6380+680=7060 km. There also a manipulation of the given equations in the examples for velocity, sqr(G*m(earth)/r))

I am probably thinking along the wrong lines, but shouldn't the shuttle be moving at the same speed needed to keep the satellite in orbit?

So that is my approach to the problem thus far - trying to find the velocity of the satellite and equating it to the velocity of the shuttle upon release. To do that, I've used the sqr(G*m(earth)/r)) equation, which gives me:

sqr((6.67*10^-11)(5.98*10^24)/(7060))=237690 m/s. This isn't even close to the given answer in the book, 7520 m/s.
 
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jmcmillian said:
The space shuttle releases a sattelite into a circular orbit 680 km above the Earth. How fast must the shuttle be moving (relative to the Earth's center) when the release occurs?I know that G (m*m(earth)/r^2) = m (v^2/r)...with r = r(earth) + h. In this equation, r=6380+680=7060 km. There also a manipulation of the given equations in the examples for velocity, sqr(G*m(earth)/r))

I am probably thinking along the wrong lines, but shouldn't the shuttle be moving at the same speed needed to keep the satellite in orbit?

So that is my approach to the problem thus far - trying to find the velocity of the satellite and equating it to the velocity of the shuttle upon release. To do that, I've used the sqr(G*m(earth)/r)) equation, which gives me:

sqr((6.67*10^-11)(5.98*10^24)/(7060))=237690 m/s. This isn't even close to the given answer in the book, 7520 m/s.
What should the units be for R...I am getting an answer very close to what your book says.

Casey
 
Last edited:
Ah yes, it should be m instead of km. Often times I get caught up in the problem and lose focus in critical things such as units. Thanks for your help.
 
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