Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @satellite73

    • one year ago
  2. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @phi

    • one year ago
  3. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  4. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @satellite73 @phi How does the above simplify to \[=-\frac{GMm}{2r}\]?

    • one year ago
  5. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix

    • one year ago
  6. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @joemath314159

    • one year ago
  7. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm soo silly...nevermind

    • one year ago
  8. wio Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  9. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    THanks!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.