Gravitation.....1

Consider an attractive central force of the form F(r) = -k/rn where k is a constant. For a stable circular orbit to exist what should be the value/values of n?

11 Answers

1
skygirl ·

one will be n=2 for sure.

1
Surbhi Agrawal ·

can u plz exlplain it!!!

1
greatvishal swami ·

n=2 only

1
Surbhi Agrawal ·

plz provide explanation!!!!!!!!!

1
rahul wadhwani ·

i think if i m right then n≥1 as for stable orbit -∂F/∂r is negative as potential energy is minimum thinking so ..

11
Anirudh Narayanan ·

EXPLANATION PLEASE? [7]

1
rahul wadhwani ·

sorry and thanx aragorn for giving those comment i had use wrong relation its answer will be n≥2 as ∂U/∂r =-F and for stable equilibrium P.E. must be negative then intgrate it then u will get if n≥2 then we will get -ve P.E. i had not use vector notatiuonof formula ok

1
°ღ•๓яυΠ·

attractive force=KQq/r^2

so one value is 2

11
Anirudh Narayanan ·

F(r) = -k/rn
What is the significance of the negative sign? Can't we combine it with k and write it as another K?

1
skygirl ·

this is a nice question... i think rahul is nearest to the explanation...

we need to kp in mind dat 'stable' thing...

but there's something more lucid...

well still thinking ....

11
Anirudh Narayanan ·

Does the negative sign indicate the direction of the force?

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