anonymous
  • anonymous
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?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
@phi
anonymous
  • anonymous
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 \[E=\frac 12mv^2-\frac{GMm}{r}\] \[E=\frac 12m\frac{GM}{r}-\frac{GMm}{r}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@satellite73 @phi How does the above simplify to \[=-\frac{GMm}{2r}\]?
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
@joemath314159
anonymous
  • anonymous
I'm soo silly...nevermind
anonymous
  • anonymous
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.
anonymous
  • anonymous
THanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.