Question 42. At what distance above and below the surface of the Earth will a body have the same weight?
Take the radius of the Earth as R.
(1) \sqrt{5}R-R
(2) \frac{\sqrt{3}R-R}{2}
(3) \frac{R}{2}
(4) \frac{\sqrt{5}R-R}{2}
Answer: (4)
EXplanation:
Let the distance from the surface be h.
Weight is proportional to acceleration due to gravity.
Acceleration due to gravity at a height h above the surface is
g_{p}=\frac{gR^{2}}{(R+h)^{2}}Acceleration due to gravity at a depth h below the surface is
g_{q}=g\left(1-\frac{h}{R}\right)For the weights to be equal, g_{p}=g_{q}
So, \frac{g}{\left(1+\frac{h}{R}\right)^{2}}=g\left(1-\frac{h}{R}\right)
Cancel g from both sides,
\frac{1}{\left(1+\frac{h}{R}\right)^{2}}=1-\frac{h}{R}Multiply both sides, \left(1-\frac{h}{R}\right)\left(1+\frac{h}{R}\right)^{2}=1
Let \frac{h}{R}=x
Then, (1+x)^{2}=1
x^{3}-x^{2}-x=0Solving, x=\frac{\sqrt{5}-1}{2}
So, h=\frac{R}{2}(\sqrt{5}-1)
Hence, the required distance is \frac{\sqrt{5}R-R}{2}.