Find the dimensions of G in the equation F-dfrac-Gm-1-m-2-r-2- - where F is the force of gravitational attraction between two bodies of point masses m-1- and m-2- - -r- is the distance of separation between them-

Question

Find the dimensions of G in the equation F-dfrac-Gm-1-m-2-r-2- - where F is the force of gravitational attraction between two bodies of point masses  m-1- and m-2- - -r- is the distance of separation between them-

Toppr Admin · Accepted Answer

The gravitational force acting between the two bodies is given by-F-Gm1m2r2G-Fr2m1m2The dimension of the force is -MLT-x2212-2-MLT-x2212-2-L2-M-M-M-x2212-1L3T-x2212-2

Find the dimensions of G in the equation $F = \frac{G m_{1} m_{2}}{r^{2}}$ ; where F is the force of gravitational attraction between two bodies of point masses $m_{1}$ and $m_{2}$ , 'r' is the distance of separation between them.

The gravitational force acting between the two bodies is given by:
$F = G \frac{m_{1} m_{2}}{r^{2}}$

$G = \frac{F r^{2}}{m_{1} m_{2}}$

The dimension of the force is $[M L T^{- 2}]$

$= \frac{[M L T^{- 2}] [L^{2}]}{[M] [M]} = M^{- 1} L^{3} T^{- 2}$

Find the dimensions of G in the equation F=Gm1m2r2 ; where F is the force of gravitational attraction between two bodies of point masses m1 and m2 , 'r' is the distance of separation between them.

The gravitational force acting between the two bodies is given by:F=Gm1m2r2G=Fr2m1m2The dimension of the force is [MLT−2]=[MLT−2][L2][M][M]=M−1L3T−2

Find the dimensions of G in the equation $F = \frac{G m_{1} m_{2}}{r^{2}}$ ; where F is the force of gravitational attraction between two bodies of point masses $m_{1}$ and $m_{2}$ , 'r' is the distance of separation between them.

The gravitational force acting between the two bodies is given by:
$F = G \frac{m_{1} m_{2}}{r^{2}}$

$G = \frac{F r^{2}}{m_{1} m_{2}}$

The dimension of the force is $[M L T^{- 2}]$

$= \frac{[M L T^{- 2}] [L^{2}]}{[M] [M]} = M^{- 1} L^{3} T^{- 2}$