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JenniferSmart1

  • 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?

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  1. JenniferSmart1
    • one year ago
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    @satellite73

  2. JenniferSmart1
    • one year ago
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    @phi

  3. JenniferSmart1
    • 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 \[E=\frac 12mv^2-\frac{GMm}{r}\] \[E=\frac 12m\frac{GM}{r}-\frac{GMm}{r}\]

  4. JenniferSmart1
    • one year ago
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    @satellite73 @phi How does the above simplify to \[=-\frac{GMm}{2r}\]?

  5. JenniferSmart1
    • one year ago
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    @zepdrix

  6. JenniferSmart1
    • one year ago
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    @joemath314159

  7. JenniferSmart1
    • one year ago
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    I'm soo silly...nevermind

  8. wio
    • one year ago
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    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.

  9. JenniferSmart1
    • one year ago
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    THanks!

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