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

Question

If the gravitational force between two objects were proportional to 1/R (and not as 1/R2 ), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to (A) R (B)  R0 (independent of R) (C) 1/R2 (D) 1/R

Tardigrade Admin · Accepted Answer

Centripetal force (F)=(m v2/R) and the gravitational force (F)=(G M m/R2)=(G M m/R) (where R2 arrow R ). Since (m v2/R)=(G M m/R), therefore v=√G M. Thus velocity v is independent of R.

Q. If the gravitational force between two objects were proportional to $1/ R$ (and not as $1/ R^{2}$ ), where $R$ is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to

$R$

$R^{0}$ (independent of $R$ )

$1/ R^{2}$

$1/ R$

Centripetal force $(F) = \frac{m v ^{2}}{R}$
and the gravitational force
$(F) = \frac{GM m}{R ^{2}} = \frac{GM m}{R}$
(where $R^{2} \to R$ ).
Since $\frac{m v ^{2}}{R} = \frac{GM m}{R}$ ,
therefore $v = GM$ .
Thus velocity $v$ is independent of $R$ .

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

R

R0 (independent of R)

1/R2

1/R

Centripetal force (F)=Rmv2​ and the gravitational force (F)=R2GMm​=RGMm​ (where R2→R ). Since Rmv2​=RGMm​, therefore v=GM​. Thus velocity v is independent of R.

Q. If the gravitational force between two objects were proportional to $1/ R$ (and not as $1/ R^{2}$ ), where $R$ is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to

$R$

$R^{0}$ (independent of $R$ )

$1/ R^{2}$

$1/ R$

Centripetal force $(F) = \frac{m v ^{2}}{R}$
and the gravitational force
$(F) = \frac{GM m}{R ^{2}} = \frac{GM m}{R}$
(where $R^{2} \to R$ ).
Since $\frac{m v ^{2}}{R} = \frac{GM m}{R}$ ,
therefore $v = GM$ .
Thus velocity $v$ is independent of $R$ .