Explanation
Ratio of time period of satellites orbiting around earth at a distance of about 9r and 4r from earth surface is : ( where r is the radius of earth)
Let $$g_e= 10 m/s $$
$$R_e =6400m$$
$$g'=?$$
hence by formula
$$g'-g_e = \dfrac{1}{2}\times \dfrac{h}{R_e}$$
thus g=-4.9 m\s
g = G M / R2
as, R is decreased by 1.5%
So, new Radius , R' = R - 0.015 R = 0.985R
So, new g, g' = G M / (0.985R) 2
= 1.0306 G M/ R2
So, change = 0.0306
% change = 3 % (approx)
In the figure it is shown that the velocity of lift is $$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ while string ins winding on the motor shaft with velocity $$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ and shaft A is moving downward with velocity $$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ with respect lift, then find out the velocity of block B
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