two celestial bodeis are seperated by some distance .if the mass of any one of the point like bodies is doubled while the mass of other is halved then how far should they be taken so that the gravitational force between them becomes one fourth
Answers
Answered by
51
F = Gm1m2/r^2
Gm1m2 = Fr^2
m1'= 2m1
m2' = m2/2
F'= F/4
r' = ?
F' = Gm1'm2'/r'^2
F/4 = G(2m1 × m2/2)/ r'^2
F/4 = Gm1m2/r'^2
r'^2 = 4Gm1m2/F
r'^2 = 4×F×r^2/F
r' = √4r^2
r' = 2r
Therefore the distance between two celestial bodies must be made twice.
Gm1m2 = Fr^2
m1'= 2m1
m2' = m2/2
F'= F/4
r' = ?
F' = Gm1'm2'/r'^2
F/4 = G(2m1 × m2/2)/ r'^2
F/4 = Gm1m2/r'^2
r'^2 = 4Gm1m2/F
r'^2 = 4×F×r^2/F
r' = √4r^2
r' = 2r
Therefore the distance between two celestial bodies must be made twice.
Diksha123:
thanku
plzzz mark as brainliest.....
kk
plzzz mark mine answer as brainliest answer
actually there was no option of brainliest answer
ohh
ya
Similar questions