1. Not finding help here? Sign up for a free 30min 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!

Escape speed in space

  1. Apr 28, 2005 #1
    QUESTION:
    A space station orbits the sun at the same distance as the earth but on the opposite side of the sun. A small probe is fired away from the station. What minimum speed (m/s) does the probe need to escape the solar system?

    MY UNDERSTANDING AND SOLUTION:
    The escape speed v from a sphere of radius R and mass M is given by the energy-conservation equation as follows: (from "Schaum's 3000 Solved Problems in Physics" book, page 101)

    (1/2)*m*v^2 = G*M*m / R where;

    M: mass of the sun (=1.98*10^30 kg)
    m: mass of the small probe
    R: Radius of the sun (=6.95*10^8 m)
    G: Universal Gravitation Constant [=6.67*10^(-11) Nm^2/kg^2]

    From the equation typed in bold above;

    v = sqrt(2*G*M / R) and using the numerical values v is found about

    616479 m/s .

    What do you think?

    Thanks.
     
  2. jcsd
  3. Apr 28, 2005 #2

    DaveC426913

    User Avatar
    Gold Member

    Six hundred kilometers per second???

    Does that sound reasonable???

    Always use your head to check your answers.



    Without examining too closely, one thing I note is that you haven't calc'd the escape velocity from Earth orbit, you've calc'ed the escape velocity from the surface of the Sun: R = 695,000km. R should be Earth's orbit.
     
  4. Apr 28, 2005 #3
    That's why I had to ask.

    Okay, let us take R as the distance between the sun and earth since the problem says "A space station orbits the sun at the same distance as the earth but on the opposite side of the sun", then v is found

    51,393.77 m/s. What about this one?

    Is this reasonable? Our professor says yes, but it is incorrect. So what?
     
  5. Apr 28, 2005 #4

    BobG

    User Avatar
    Science Advisor
    Homework Helper

    Are you sure your numbers were right? You're in the ballpark, but definitely wrong.

    The heliocentric gravitational constant (GM) is 1.32712442076 x 10^11 km

    One astronomical unit (average radius of the Earth's orbit) is 1.49597871 x 10^8 km

    Just eyeballing it tells you your escape velocity will be a little less than the square root of 2000, which puts you in 40's of km/sec.
     
  6. Apr 13, 2010 #5
    your equation is right,

    vescape = [2*G*M/R].5

    your constants must be off

    be sure to use:
    G = 6.67x10-11
    M = (mass of the sun) = 1.99x1030 kg
    R = (mean disntance from the earth to the sun) = 1.5x1011 meters

    that should give you the right answer
     
  7. Apr 13, 2010 #6
    by the way, the answer should be 4.21x104 m/s
     
  8. Apr 13, 2010 #7

    DaveC426913

    User Avatar
    Gold Member

    That's great but if he's still struggling with this after five years, he's got bigger problems... :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Escape speed in space
  1. Escape Speed (Replies: 1)

  2. Escape speed (Replies: 2)

  3. Escape speed (Replies: 2)

  4. Escape speed (Replies: 3)

  5. Escape Speed (Replies: 2)

Loading...