Gravity/Planetary Forces Question - High School physics question

AI Thread Summary
To find the distance where an object is perfectly balanced between the Earth and the Sun, the gravitational forces from both bodies must be equal. The relevant equation involves the gravitational force formula, where the mass of the object cancels out, simplifying the problem. The distance from the Earth is denoted as 'r', while the distance from the Sun is 'R - r', where 'R' is the total distance between the Earth and the Sun. By equating the gravitational forces from both celestial bodies, one can derive a formula for 'r' based on the known masses and distance. This approach effectively leads to a solution for the object's position in the gravitational field.
closertolost
Messages
14
Reaction score
0

Homework Statement


Find the distance an object needs to be in between the sun and the Earth for it to be perfectly balanced (not moving)

Homework Equations


mass of Earth = 5.98e24 kg
mass of sun=1.991e30 kg
distance between sun and Earth = 1.479e11 m
Fg=Gm1m2/r²

The Attempt at a Solution


well, i tried making the Fg zero for two equations - one with Fg of the object to the sun, and one with Fg of the object to the earth..but it got messed up..can someone help me?

can i sub in the mass of the object as 1kg?
 
Physics news on Phys.org
Hint: The point at which the object won't accelerate is the point at which the gravitational forces on the object are *equal* in magnitude (and opposite in direction).

No, you don't need to substitute in a numerical value for the mass of the object, because it should cancel from both sides of the equation (meaning that the point at which the forces balance doesn't depend upon how heavy the object is).
 
so, do i put
Fg=(6.67e-11 x 5.97e24)/d²
Fg=(6.67e-11kg x 1.991e30kg)/1.479e11m-d² <---because d is the distance to the object, and 1.479e11 is the distance from sun to Earth (?)
 
oh, and then cram the equations together :P
 
closertolost

When I hinted that the gravitational forces on the object due to Earth and the sun were equal, I meant for you to actually *equate* them. So, using the subscripts E and S for Earth and sun respectively, we have:

FE = FS

GmME/r2 = GmMS/(R-r)2

You'll notice I've done something interesting here with the distances. I've decided to call the distance between the Earth and the sun "R." So, if the object is a distance r from Earth, then its distance from the sun must necessarily be R-r (since the problem states that the object lies along a straight line connecting the two celestial bodies). If you're still not sure, draw a diagram. EDIT: I see that you already figured this out, nice work.

You'll also notice that m, the mass of the "test object" cancels from both sides of the equation, and so does the gravitational constant, G. This leaves you with an equation for r (the thing you are trying to calculate) in terms of three known quantities. The three known quantities are the masses of the Earth and sun, and the Earth-sun distance, R.
 
ahah thanks very much! i have it now :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top