Expression for the gravitational potential Vgrav

AI Thread Summary
To derive the gravitational potential Vgrav at a position x on the x-axis due to two equal masses fixed at x=0 and x=x0, start with the gravitational potential formula Vgrav(r) = -GM/r for each mass. Calculate the potential from each mass separately, considering their distances from point x, and then sum these potentials to find the total gravitational potential at that point. The first mass contributes V1 = -Gm/(x) and the second mass contributes V2 = -Gm/(x - x0). The final expression for the total gravitational potential is Vgrav = V1 + V2. Understanding this process clarifies how to approach problems involving multiple masses and their combined effects on gravitational potential.
c_m
Messages
9
Reaction score
0

Homework Statement



In a system where other gravitational influences can be discounted, two particles of equall mass m, are fixed at positions x= 0 and x= x0 on the x-axis.

1) derive an expression for the gravitational potential Vgrav at a general position x on the x-axis.?



Homework Equations



The only equations i can seem to find are:

A) for a test mass m at a distance r from another mass M
Vgrav(r) = (1/m)Egrav = (1/m)(-GmM/r) = -GM/r

B) V(r) = Q/ 4pie e0 r
Where e0 is the permittivity of free space



The Attempt at a Solution



Well to derive an expression usually means i have to combine and rearrange two separate expressions, but i do not know where to begin, maybe i don't need either of these expressions? could somebody please help me to understand what is goin on here? and prehaps where to start looking?
 
Physics news on Phys.org
Equation A you have there is the gravitational potential. Equation B is the electrostatic potential so you won't need that. So you have two masses along the x-axis and you want to know the potential along the axis dues to those two masses. How do you think you should proceed from here?
 
Thankyou for relpying

I have actually sent my work now but i did not really do this one, so i would still like to go through it to see what i should have done.

So i do need equation A then? but not B. do i need to find another expression now then for the potential? I just didnt know where to begin.
 
When there is more than one mass involved you can sum the potentials.
 
you mean like, V = G m1 m2 / r? rather than V = Gm/r?
 
No. Work out the potential using equation A due to one mass, then the other and then add them together to find the potential due to both.
 
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