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Answer is - v = [Gm (1/2R - 1/r)]

^{1/2}[answer := 5.62 * 1024 kg ]

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Three objects each of mass 'm' are placed at the corners of an equilateral triangle of side 'l' . What is the gravitational force acting on each mass due to other masses is

$\text{1)}\frac{{\text{Gm}}^{2}}{\sqrt{3{\text{l}}^{2}}}\phantom{\rule{0ex}{0ex}}\text{2)}\frac{\sqrt{3}{\text{Gm}}^{2}}{2{\text{l}}^{2}}\phantom{\rule{0ex}{0ex}}\text{3)}\frac{\sqrt{3}{\text{Gm}}^{2}}{{\text{l}}^{2}}\phantom{\rule{0ex}{0ex}}\text{4)}\frac{{\text{Gm}}^{2}}{{\text{l}}^{2}}$

a)T4/3

b)T2/3

c)T1/2

d)T

give solution?

Q. When two identical sphere are kept in contact, the gravitational force between them is F. If two sphere of same material but with

twice the radius are radius are kept in contact, then the gravitational force between them would be

1) F/4

2) 4F

3) F/16

4) 16F

a) the gravitational potential at A

b) the gravitational field at A

c) the velocity with which the particle strikes the centre O of the sphere( neglect earth's gravity)

habove the surface of the Earth. Thenhis equal toB)R/2

C)2R/3

D)3R/2

plzsolve this sir using g'= g ( 1 - 2h/R) and not g' = g R^2/ (R+h)^2

we should get the same ans but i am not getting the same ans?help

R2

1) $1\times {10}^{4}$m

2) $0.32\times 10{m}^{4}$

3) $5\times {10}^{4}m$

4) $0.032\times {10}^{4}m$