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anonymous
 3 years ago
When asked to calculate the energy required to launch a satellite into orbit, do I have to solve for E (E=K+U) or just potential U?
anonymous
 3 years ago
When asked to calculate the energy required to launch a satellite into orbit, do I have to solve for E (E=K+U) or just potential U?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think it's E=K+U \[K=\frac 12 mv^2\] \[U=\frac{GMm}{r}\] \[F=ma\] \[a=\frac{v^2}r\] \[F=m\frac{v^2}{r}\] \[F=\frac{GMm}{r^2}\] \[\frac{GMm}{r^2}=m\frac{v^2}{r}\] \[\frac{GM\cancel m}{r^2}=\cancel m\frac{v^2}{r}\] \[\frac{GM}{r}=v^2\] E=K+U \[E=\frac 12mv^2\frac{GMm}{r}\] \[E=\frac 12m\frac{GM}{r}\frac{GMm}{r}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 @phi How does the above simplify to \[=\frac{GMm}{2r}\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm soo silly...nevermind

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Orbits are elliptical so the potential energy is going to be changing, likewise kinetic energy is going to change. Clearly the amount of energy it took is constant.
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