1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating time of jump in space

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Imagine that you are helping NASA plan a manned expedition to an asteroid. The asteroid has a radius of 20km, and a surface gravity of 0.034 m s-2.

    NASA are concerned that the gravity on this asteroid is so small that if an astronaut inadvertently jumps, it would take too long for them to float back down again. They have asked you to estimate how much time it would take a jumping astronaut to return to the surface.

    You asked some astronauts to see how high they could jump in their space suits, here on Earth. It turns out that they could raise their centres of mass by 0.30 m.

    If they did a similarly powerful jump on the asteroid, how long would it take them to come back down? Type your answer, in seconds to at least one decimal place, in the box below. Do not type units.

    You may assume that the gravity on the surface of the Earth is 9.8 ms-2, and that the height of their jump on the asteroid is much less than the radius of the asteroid.

    2. Relevant equations

    3. The attempt at a solution

    i thought you could use the kinematics equation d=v0t + 1/2at^2
    since vertical velocity will be 0 at maximum height the equation would be d=1/2at^2

    i thought you could use this equation to get the time, which would give you 4.2s but thats not the answer
    since the jump on earth is 0.3 m does that mean the jump in space would be the same height? cause that is what i was assuming
    Last edited: Mar 1, 2014
  2. jcsd
  3. Mar 2, 2014 #2


    User Avatar
    2017 Award

    Staff: Mentor

    No. You can assume that the initial velocity (and therefore the kinetic energy the astronaut can put in the jump) is the same - this is not exactly true, but a good approximation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted