Suppose that in a parallel Universe the gravitational force between two point particles, instead of falling with the square of the distance, falls with the power 3/2. That is, the modulus of the attractive force between the particles of masses m1 and m2 is given in this Universe by: (see image), where K is the universal constant. l) I know that the formula for the

icon
Related questions
Question

Suppose that in a parallel Universe the gravitational force between two point particles, instead of falling
with the square of the distance, falls with the power 3/2. That is, the modulus of the attractive force between the
particles of masses m1 and m2 is given in this Universe by: (see image), where K is the universal constant.

l) I know that the formula for the gravitational potentital energy is : -(GMm/r)

Il) I know that the formula for the gravitational force is: GMm/r2

QUESTION: What's the formula for the gravitational potencial energy? Provide the answer explaining the question below

 

Looking at the formulas, it seems that the only difference between the two formulas is the 'r2', so, if the the r2 in the force is r3/2, r should be r3/4but, according to the resolution(see 2nd image) it is r1/2. Why?

DISCLAIMER: Explain why "r" cannot be r3/4, even though it apparently matches with the formula. Also, explain if there's a way to get to the gravitational potential formula without using integrals, that is, how to know that "r" must be r1/2 and why must be a "2" multiplying the numerator.

No caso, temos F =
F
kmima. Portanto,
3/2
U(r)
dr
=
2km₁m₂
r1/2
- [Fdr.
U=-
+ constante
Transcribed Image Text:No caso, temos F = F kmima. Portanto, 3/2 U(r) dr = 2km₁m₂ r1/2 - [Fdr. U=- + constante
No caso, temos F =
F
kmima. Portanto,
3/2
U(r)
dr
=
2km₁m₂
r1/2
- [Fdr.
U=-
+ constante
Transcribed Image Text:No caso, temos F = F kmima. Portanto, 3/2 U(r) dr = 2km₁m₂ r1/2 - [Fdr. U=- + constante
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer