# Gravitational Force On A Spaceship

1. Oct 9, 2005

### Lucretius

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. Oct 9, 2005

### Skippy

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

3. Oct 9, 2005

### Lucretius

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.

4. Oct 9, 2005

### 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

5. Oct 9, 2005

### Janus

Staff Emeritus
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.

6. Oct 9, 2005

### Lucretius

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…

7. Oct 9, 2005

### Pyrrhus

Basicly you got by simmetry as it has been stated

$$\sum F_{y} = 0$$

$$\sum F_{x} = 2 G \frac{m_{asteroid}m_{spaceship}}{D^2 +d^2} \frac{D}{\sqrt{D^2 + d^2}}$$

8. Oct 9, 2005

### Lucretius

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.

9. Oct 9, 2005

### Staff: Mentor

You find the x-component of the force the same way you find the x-component of any vector: $F_x = F \cos \theta$, 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.)

10. Oct 9, 2005

### Lucretius

Doc, I could kiss you!