Gravitational Potential at a Midpoint: What is the Solution?

AI Thread Summary
The discussion focuses on calculating gravitational potential and field strength at a midpoint between two equal masses. It establishes that the gravitational field strength cancels to zero due to equal and opposite forces. For gravitational potential, the participant initially misinterprets the formula but later realizes that the correct answer involves summing the potentials from both masses, leading to a potential of -4GM/r. The ambiguity in defining gravitational potential at infinity is noted, emphasizing that potential can be adjusted by a constant without affecting physics. Ultimately, the potential at the midpoint cannot be zero, ruling out one of the answer choices.
pyman999
Messages
7
Reaction score
0

Homework Statement


9500186bdc.jpg


Homework Equations


g = G*M / r^2, where g is the gravitational field strength, G is the gravitational constant, M is the mass of the attracting body, r is the distance from the center of mass of the body.
V = -G*M / r, where V is the gravitational potential.

The Attempt at a Solution


For the gravitational field strength, as P is at a midpoint, we can say that r is 1/2*r, and as the masses are also equal, G*M / 1/2*r^2 = G*M / 1/2*r^2 (as they are equal, and in opposite directions), and they will simply cancel to 0. Therefore, it can't be A or B.

For gravitational potential, again, the masses are equal, and r is 1/2*r, so -GM / 1/2*r = -GM / 1/2*r, they will again cancel to 0. However, the answer is apparently C, where gravitational potential is -4G*M / r? I can't see why.
 
Physics news on Phys.org
In V = -G*M / r there are no vectors, only scalars. r is a distance in this formula.

(dividing by a vector is awkward...)
 
BvU said:
In V = -G*M / r there are no vectors, only scalars. r is a distance in this formula.

(dividing by a vector is awkward...)
I see, so then you're left with -2GM / r + -2GM / r, my mistake.
 
The problem statement should clarify that the potential "at infinity" is defined as zero. Otherwise the question is ambiguous - you can always add a constant value to the potential without changing physics.
Using this, you can rule out answer (D) simply from the fact that an object there would need energy to escape, therefore the potential there cannot be zero.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top