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Gravitational field

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A spaceship makes a trip from the earth to the moon 380,000 km awawy. at what point in the trip will the gravitational field be zero ? the mass of the moon is about 1/81 that of the earth


    2. Relevant equations

    Fg = Gm1m2 / r^2

    3. The attempt at a solution

    ummmmmm i know that the distance is 380,000,000m
    the mass of the moon is 1/81 of the moon the mass
    of the earth, but i don't know any equation which
    will give me a gravitational field of 0 and i don't know
    the mass of the spaceship......... what do i do??
     
  2. jcsd
  3. Apr 8, 2008 #2
    Try either using a test mass for Newton's Law of Universal Gravitation to find a location where the net force is zero, or try summing the gravitational potentials such that they be zero.

    [tex]\phi=\frac{GM}{R}[/tex]
     
  4. Apr 8, 2008 #3
    what would the M be in this case ? the earth ? the spaceship? or the moon?
     
  5. Apr 8, 2008 #4
    M is the mass of the object that is creating the gravitational potential. so if you want to calculate the potential around the earth, use the the mass of the earth. if you want to calculate the potential around the moon, use the mass of the moon.
     
  6. Apr 8, 2008 #5
    okay i got GM / R = 12958.60949 ............ but i still don't understand where i'm trying
    to get to
     
  7. Apr 8, 2008 #6
    Calculate where the gravitational force caused by the moon is equal to the gravitational force caused by the earth.
     
  8. Apr 8, 2008 #7
    so what ur saying is
    GMmoon/r = Gmearth / r ???
     
  9. Apr 8, 2008 #8
    GM/r is not the equation for force or field.
     
  10. Apr 8, 2008 #9
    yargh i'm so lost ..................... lets start from square 1
     
  11. Apr 8, 2008 #10
    [tex]F=\frac{GmM}{r^2}[/tex]

    You have two forces, one from the moon and one from the earth; equate them.
     
  12. Apr 8, 2008 #11
    Gravitational field is Fg/m...its technically dependant on the source mass and independent of a test mass. Now, basically the complexity of the situation is that you're dealing with two gravitational fields: one from the moon and one from the Earth. I'm thinking the gravitational field vectors points towards the source mass, so the field vectors of the Earth and moon should be relatively in opposite directions at certain points. Those points are where you may have a field of 0N/kg. All you have to do is set the field equations for the Earth and Moon equal to each other at some point a distance relative to the moon and Earth.
     
  13. Apr 8, 2008 #12
    okay what i did was

    [tex]{G*Mmoon*Mship}/{R^2}[/tex] = [tex]{G*Mearth*Mship}/{R^2}[/tex]

    I cancelled the Mships and then I solved for one of the R of the earth
    because the distance from the earth to the moon is 380,000 km ....
    i got 3.42 x 10 ^ 6 km .............. does that sound right??
     
  14. Apr 8, 2008 #13
    okay what i did was

    [tex]{G*Mmoon*Mship}/{R^2}[/tex] = [tex]{G*Mearth*Mship}/{R^2}[/tex]

    I cancelled the Mships and then I solved for one of the R of the earth
    because the distance from the earth to the moon is 380,000 km ....
    i got 3.42 x 10 ^ 6 km .............. does that sound right??
     
  15. Apr 8, 2008 #14
    Think about your answer for a moment. 3.42 x 10^6 km is almost halfway to the planet Jupiter.

    Your R's are obviously not going to be the same for both. You are not at the moon when the net force is zero, so R will not be 380,000 km for one of them.
     
  16. Apr 8, 2008 #15
    ok nvm i got it i got 3.457 x 10^5 km
     
  17. Apr 8, 2008 #16
    My answer is somewhat off from yours, but that could be a rounding error. I won't know unless you show work.
     
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