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!

Need help with Gravitational Field Strength question

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A satellite orbites Earth at a distance of 3rEARTH, above Earth's surface. What is the magnitude of gravitational field strength at the point where the satellite is?


    2. Relevant equations
    Fg= Gm1m2/d^2 (m1=mass 1, m2=mass2)
    F1/F2 = D2^2/D2^2 (D1 is a variable, and so is D2, so it represents Distance 1 and Distance 2)


    3. The attempt at a solution
    I don't understand how it works? You have the equation for the radius of Earth, 6400km, but not much else other than the G which is 6.67x10^-11. I'm not sure how to tackle this, and it's confusing me a lot. I tried applying Fg= Gm1m2/d^2, but there is no mass on them.


    P.s on a secondary note, if I am doing a question where it compares 2 masses that have an attraction (i.e 36N, how do you find the distances/mass?)
     
  2. jcsd
  3. Nov 2, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The gravitational attraction between masses m1 and m2, d distance apart, is F=Gm1m2/d2. If the objects are spheres d means the distance between their centres.

    The gravitational field strength means the force of gravity exerted on unit mass. At the surface of the Earth it is g=9.8 m/s2:

    9.8=GM(Earth)/R2, where R is the radius of the Earth.

    The gravitational field strength at distance d from the centre of Earth is g'=GM(Earth) /d2.
    If you divide the equations in bold you get

    g'/9.8=R2/d2.

    The satellite is 3R above the surface of Earth. How far is it from the centre of Earth?

    ehild
     
  4. Nov 2, 2012 #3
    The gravity of the Earth at some distance is inversely proportional to the square of the distance from its center. At the surface, this is g. That, and the distance of the satellite from the Earth, should be enough.
     
  5. Nov 3, 2012 #4
    I understand the question now, thanks for the help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Need help with Gravitational Field Strength question
Loading...