Calculating Force on Spaceship: G, m1, m2, r^2

[drkidd22]

Homework Statement

If D = 2323 meters and d = 1234 meters in the figure, what is the magnitude of the force on the spaceship to the nearest MN? Consider all three to be point masses.

Homework Equations

F=(Gm1m2)/r^2

The Attempt at a Solution

For this problem I have tried to use the above equation and have not gotten it correct yet. The distance from the ship to either asteroid is 2630m. MN I guess is for MegaN?

any help will be appreciated.
Thanks
Jose

 

[Dick]

Yes, I think it's mega Newtons. You need to use some trig here. Draw a force diagram and split it into horizontal and vertical components. The vertical ones on the spaceship cancel. The horizontal ones add.

 

[drkidd22]

I still can't get this thing right. Do I have to find the angles first then the difference of the forces?

 

[Dick]

I still can't get this thing right. Do I have to find the angles first then the difference of the forces?

What angles are you talking about? The gravitational forces act along a vector between the ship and the two asteroids, right? You can either figure out what they both are as vectors and then add the two vectors or you can realize that you only need the component of the vectors along the line you labelled 'D'. You have done free body diagrams and split forces into components and stuff like that, I hope.

 

[drkidd22]

This is what I just did.

C^2 = 2323^2 + 1234^2
C = 2360
The hypotenous from ship to asteroid on right top is 2630

[(10^-10)(3.5*10^18)(2.5*10^7)]/(2630)^2

I still think I'm missing something here. Don't know if I did this diagram correct. Don't have a scanner to upload it.

 

[Dick]

This is what I just did.

C^2 = 2323^2 + 1234^2
C = 2360
The hypotenous from ship to asteroid on right top is 2630

[(10^-10)(3.5*10^18)(2.5*10^7)]/(2630)^2

I still think I'm missing something here. Don't know if I did this diagram correct. Don't have a scanner to upload it.

The distance is fine. The force equation is about right, (though if you want to get within one mega Newton you'd better use a better approximation for G). The BIG TROUBLE here is that that force should be a VECTOR. What you have there is just the length of the vector. Draw a arrow between the ship and the asteroid and label it with that number. That's one of your force vectors, the other one points at the other asteroid. To add them you have to add vectors! I keep saying this. Do you know what a vector is?

 

[drkidd22]

Isn't that what I just did with the above equation. Find the Force of that vector?[(6.673*10^-11)(3.5*10^18)(2.5*10^7)]/(2630)^2 << that would be the force on the top vector right?

Now I find the one in the bottom and and the two force vectors??

Both asteroids are at the same distance from the ship, they both have the same mass, so I would just need to multiply the force of the found vector *2 and I get 1688MN

 

[Dick]

Isn't that what I just did with the above equation. Find the Force of that vector?[(6.673*10^-11)(3.5*10^18)(2.5*10^7)]/(2630)^2 << that would be the force on the top vector right?

Now I find the one in the bottom and and the two force vectors??

Both asteroids are at the same distance from the ship, they both have the same mass, so I would just need to multiply the force of the found vector *2 and I get 1688MN

You are getting the magnitude of each force correctly. But you can only add the magnitudes of the two forces if they point in EXACTLY the same direction. For example, if they were pointed in opposite directions, they would cancel giving you zero. You HAVE to do a vector addition. You will get a magnitude of the total force somewhat less than twice the magnitude of a single force.

 

[Cryphonus]

I didnt really get what you are asking here maybe you can or someone can clarify it for me, but when i look at the diagram i see that you can ignore vertical forces on the spaceship since the sum of them is zero.So you can calculate the horizontal force on the spaceship with that formula you mentioned...

 

[drkidd22]

ok so my horizontal vector will be (2323,1234) + (2323, -1234) = (4646,0). Right? then I use that in the formula and I get ~270MN. I still think I either don't get it or I don't know what I'm doing, but I been trying really hard. If someone could walk me thru this I'll appreciate it. Not giving up until I get it.

Thanks

 

[Dick]

I can't fault you for not trying and don't give up. But it really seems like don't know how to work with vectors. Ok, I'll walk you through it. You know each force vector has a magnitude of about 844MN, right? Stop me when you don't get something. Each vector makes an angle with the horizontal. Split each vector into horizontal and vertical parts whose sum is the whole vector. You can find the horizontal parts by multiplying the magnitude by the cosine of the angle, and you can find the vertical parts by multiplying by the sine of the angle. That's the trig part. Have you ever seen this before? You don't need to find the vertical parts because you can see from the diagram they point in opposite directions and will cancel. Are you still with me? You do need to find the horizontal magnitudes because they point in the same direction and will add. So the total force is directed towards a point in between the two asteroids. And that magnitude is 2*cos(angle)*844MN. To find the angle you can use trig again and realize that the cos(angle)=adjacent side/hypotenuse of the triangle (or you could use tangent) so cos(angle)=2323/2630, you already worked the 2630 out before. Is this making any sense at all? And you may need to check how some of these figures round off to make sure you are within 1MN of the answer. That's the best I can do for a walk through.

 

[drkidd22]

That's what I mentioned above, that I needed to use the angles/find them. You said what angles so I though there was a different way of doing it. But let me get on it right now and I will get back at you.

Thanks

 

[drkidd22]

I think I got it right now. All the confusion that had me puzzled was when you said what angles. Think I might have miss-understood that, but it clicked with your last post.

thanks