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JenniferSmart1
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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?
 one year ago
 one year ago
JenniferSmart1 Group Title
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?
 one year ago
 one year ago

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@satellite73
 one year ago

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@phi
 one year ago

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I 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 \[E=\frac 12mv^2\frac{GMm}{r}\] \[E=\frac 12m\frac{GM}{r}\frac{GMm}{r}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 @phi How does the above simplify to \[=\frac{GMm}{2r}\]?
 one year ago

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@zepdrix
 one year ago

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@joemath314159
 one year ago

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I'm soo silly...nevermind
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Orbits 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.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
THanks!
 one year ago
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