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Gravitational Force On A Spaceship

  1. Oct 9, 2005 #1

    This is the problem I am working on. I do not know how I am supposed to find the force exerted on the spaceship. I figured I would find the gravitational attraction of the asteroids on eachother, and then use that combined force, but then I get stuck. I also don't know what MN stands for? Some form of a Newton? I just need a push in the right direction.
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Oct 9, 2005 #2
    Since the asteroids are the same mass and distances away, the Y-component of their attraction on the spaceship will cancel out. So just the X-component of their force will matter. Find how much force one of the asteroids exerts on the space ship in the x direction to start off.

    No idea on MN
  4. Oct 9, 2005 #3
    Okay, I must have tried about fifty things, and I get the wrong answer every time. If the y-components cancel out, the spaceship does not move upwards at all, correct? I found the force exerted by one asteroid on the spaceship. I thought I would have to find the hypotenus of a triangle, but since the spaceship doesn't move in the y direction at all… What use is the d-value? (not the big D, the small d). I'm still as confused as ever.
  5. Oct 9, 2005 #4

    Doc Al

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    Staff: Mentor

    Each asteroid exerts a gravitational force on the ship. Calculate that force. (The force depends on D and d.) If you take the x-axis as being in the direction the ship is moving, then, as Skippy says, the y-components of the force will cancel out. So find the x-component of the force and double it.

    MN = mega Newton
  6. Oct 9, 2005 #5


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    You need the d value along with the D value to find the force acting on the ship due to one asteroid and then you need it again to find how much of that is the x component.
  7. Oct 9, 2005 #6
    I had to take a break I was getting pretty mad at this problem — and I still cannot figure it out... I did the following…

    (these numbers are different than the ones in the picture, D=3803m, and d=1426m)

    I first wanted to make a triangle, so 3803^2 + 1426^2, then I took the square root of that answer to get 4061 for my resultant force.

    Then I filled in the equation Gm1m2/r^2. The answer I get is 3.5389x10^8.

    I figured if I wanted to find the total force on the spaceship, I would have to include the other force provided by the other asteroid. Because they are evenly spaced apart, and have the same mass, I know that the y-component, and the hypotenuse of my force triangle cancel out — and that the x-value is doubled.

    But, how do I get what the x-component of my force triangle is? I can't just plug in Gm1m2/r^2 again…

    Sorry I am not understanding this problem at all, it is frustrating me beyong belief…
  8. Oct 9, 2005 #7


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    Basicly you got by simmetry as it has been stated

    [tex] \sum F_{y} = 0 [/tex]

    [tex] \sum F_{x} = 2 G \frac{m_{asteroid}m_{spaceship}}{D^2 +d^2} \frac{D}{\sqrt{D^2 + d^2}} [/tex]
  9. Oct 9, 2005 #8
    I think I am just going to call it quits on this problem. Can't find a way to solve it, and it's annoying me too much.
  10. Oct 9, 2005 #9

    Doc Al

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    You find the x-component of the force the same way you find the x-component of any vector: [itex]F_x = F \cos \theta[/itex], where the angle is measured from the x-axis. You can find the cos(theta) by using the properties of the right triangle. (Look carefully at what Cyclovenom posted.)
  11. Oct 9, 2005 #10
    Doc, I could kiss you!
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