Moon Earth and Satellite gravitation

AI Thread Summary
The discussion revolves around calculating the net gravitational force on a 1030 kg spacecraft positioned at the vertices of an equilateral triangle with the Earth and Moon. The gravitational force equation is applied, but the user struggles with obtaining the correct answer, which is stated to be 2.8 N. Suggestions are made to simplify the coordinate system by aligning one of the forces along an axis to reduce complexity. The user realizes that they were not squaring the sine and cosine values in their calculations, which led to incorrect results. Ultimately, the conversation highlights the importance of careful mathematical handling and the value of collaborative problem-solving.
Frozen Light
Messages
14
Reaction score
0

Homework Statement


At a certain instant, the earth, the moon, and a stationary 1030kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84×10^5km in length.

Find the magnitude of the net gravitational force exerted on the spacecraft by the Earth and moon.

Homework Equations


GMm/r^2 = Fg

r = 3.84*10^5 km
Me = 5.98*10^24 kg
Mm = 7.35*10^22 kg
G = 6.67*10^-11
Ms = 1030kg

The Attempt at a Solution



http://imgur.com/p8OBdfd
I think it's probably a simple error somewhere but I've gotten about 3-4 different answers all which are not F=2.8N - I've tried calculating it at different stages in the algebra - are the angles somehow not based off 60*? I don't know what it's missing.
 
Last edited:
Physics news on Phys.org
Frozen Light said:

Homework Statement


At a certain instant, the earth, the moon, and a stationary 1030kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84×105km in length.
Is there a question to go along with this statement?
 
OH, sorry.

Find the magnitude of the net gravitational force exerted on the spacecraft by the Earth and moon. - edited it it in the first post too

edit: also that it's not 3.84×105km - it's 3.84 x 10^5km

edit: in the image I accidentally dropped the squares around [[ROOT][Cos60(Me - Mm)^2 + Sin60(Me+Mm)^2] but added them in my calculation
 
Last edited:
You're probably making your life more difficult than you have to by placing the satellite at the top of the triangle and making your coordinate system pass through it. Instead, try placing the satellite at the origin and the other two bodies at the other vertexes like this:

attachment.php?attachmentid=65916&stc=1&d=1390412580.gif


This way only one of the forces (FSM) will have two components.

I'd expect the net force magnitude to be just a few Newtons.
 

Attachments

  • Fig1.gif
    Fig1.gif
    1.3 KB · Views: 1,957
The answer is 2.8N - I'm trying to match my answer to that one.
But I don't really understand what you mean.

Do you mean just line one of the sides with an axis?

(Triangle)
Me
M mS

So that there is no Xdir force from Me?
 
The idea is to choose the axes so that one of the forces lies entirely along an axis so it has only one component (the other being zero). Then you only need concern yourself with the individual components of the other force.

Looking at your solution, it appears that you aren't squaring the sin and cos values when you take the sum of squares of the components.
 
Yes, that was it. Thank you.

I'm not sure how I didn't see that after spending that much time looking at it haha.
 
Frozen Light said:
Yes, that was it. Thank you.

I'm not sure how I didn't see that after spending that much time looking at it haha.

The most insidious errors are the ones that hidden in plain sight :smile: It often just takes a fresh pair of eyes to spot them.
 
Back
Top