Orbital speed Problem.

Classical Mechanics Level 2

If the gravitational force between two objects were proportional to 1/R (and not as 1/R2) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

1/R^2 1/R R^0 R

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Ishita Bhatia by Ishita Bhatia , 23, India

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2 solutions

Ishita Bhatia
Feb 5, 2015

Given that Gravitational force is proportional to 1/R

Hence Gravitational force=GMm/R

Formula for centripetal force=mv2/R

From above equation we get mv2/R=GMm/R ∴ v2=GM

v ∝R0

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Thank you so much..owsm

Masood Ansari - 4 years, 4 months ago

Though its not possible- being F = G M m R F=\frac{GMm}{R} Let's see what if it was real!

Then the acceleration- a = v 2 R a=\frac{v^2}{R}

Again if Force, F was proportional to 1 R \frac{1}{R} , Then a = G M R a=\frac{GM}{R} v 2 R = G M R \frac{v^2}{R}=\frac{GM}{R} → v 2 = G M \rightarrow v^2=GM As there's no R in the equation, V is not dependent to R .

So that's what you can say V is proportional to R 0 = 1 R^0=1 !

Its just imagination, not possible at all!

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