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!

Gravitational Force On A Spaceship

  1. Oct 9, 2005 #1
    [​IMG]

    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.
     
  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

    User Avatar

    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

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    Pyrrhus

    User Avatar
    Homework Helper

    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

    User Avatar

    Staff: Mentor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gravitational Force On A Spaceship
  1. Gravitational force (Replies: 2)

  2. Gravitational force? (Replies: 2)

Loading...