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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
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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JenniferSmart1Best ResponseYou've already chosen the best response.0
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

JenniferSmart1Best ResponseYou've already chosen the best response.0
@satellite73 @phi How does the above simplify to \[=\frac{GMm}{2r}\]?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I'm soo silly...nevermind
 one year ago

wioBest 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
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