Question

if the gravitational force between two objects is proportional to 1/R (and not to 1/R^2) where R is the separation between them, then particle in circular orbit under such force would have its orbital speed v proportional to?
OPTIONS
a)1/R^2
b)R
C)R^1
d)1/R

Answer 1

Answer on Question #43482, Physics, Mechanics | Kinematics | Dynamics

if the gravitational force between two objects is proportional to $1 / R$ (and not to $1 / R^{\wedge}2$ ) where $R$ is the separation between them, then particle in circular orbit under such force would have its orbital speed $v$ proportional to?

OPTIONS

a)1/R^2

b)R

C)R^1

d)1/R

Solution.

From Newton's $2^{\text{nd}}$ law for the particle:

$$F _ {g r a v} = m a _ {C}$$

Where $a_{C}$ is centripetal acceleration of particle.

From condition:

$$F _ {g r a v} = \frac {\alpha}{R}$$

So:

$$\frac {\alpha}{R} = m a _ {C} = m \frac {V ^ {2}}{R} \Rightarrow V = \sqrt {\frac {\alpha}{m}} = \sqrt {\frac {\alpha}{m}} \cdot R ^ {0}$$

Answer: $V \sim R^0$

no right answer among proposed options.

Options b) and c) looks the same. Check the condition.

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