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Question: Compare the weight of a \(5kg\) body \(10km\) above and \(10km\) below the surface of earth. Given: ...

Compare the weight of a 5kg5kg body 10km10km above and 10km10km below the surface of earth. Given: radius of earth =6400km=6400km

Explanation

Solution

Here we need to apply the formulas for change in acceleration due to gravity in a height and depth from the surface of the earth. Then we will compare the weights in both the cases. In both the cases the value of acceleration due to gravity is less than that on the surface of the earth.

Formula used: g=g(12hRe),g=g(1dRe){g}'=g(1-\dfrac{2h}{{{R}_{e}}}),{g}''=g(1-\dfrac{d}{{{R}_{e}}})

Complete step by step answer:
If gg and g{g}' are the value of acceleration due to gravity on the surface of the earth and at a heighthh respectively and Re{{R}_{e}} be the radius of the earth then
g=g(1+hRe)2{g}'=g{{(1+\dfrac{h}{{{R}_{e}}})}^{-2}}Now if hh is negligible compared to Re{{R}_{e}} from binomial theorem we can write
g=g(12hRe){g}'=g(1-\dfrac{2h}{{{R}_{e}}}), Now if W{W}' is the weight of a body a height hh ,it is given by
W=mg(12hRe)..........(1){W}'=mg(1-\dfrac{2h}{{{R}_{e}}})..........(1)
Now if gg and g{g}'' are the value of acceleration due to gravity on the surface of the earth and at a depthdd respectively and Re{{R}_{e}} be the radius of the earth then
g=g(1dRe){g}''=g(1-\dfrac{d}{{{R}_{e}}}), Now if W{W}'' is the weight of a at a depth hh,then it is given by
W=mg(1dRe)..........(2){W}''=mg(1-\dfrac{d}{{{R}_{e}}})..........(2)
Now dividing (1)(1) by(2)(2) we have
WW=12hRe1dRe=Re2hRed\dfrac{{{W}'}}{{{W}''}}=\dfrac{1-\dfrac{2h}{{{R}_{e}}}}{1-\dfrac{d}{{{R}_{e}}}}=\dfrac{{{R}_{e}}-2h}{{{R}_{e}}-d}
Now putting the values of all the quantities we get
WW=64002×10640010=638639 orW:W=638:639 \begin{aligned} & \dfrac{{{W}'}}{{{W}''}}=\dfrac{6400-2\times 10}{6400-10}=\dfrac{638}{639} \\\ & or{W}':{W}''=638:639 \\\ \end{aligned}

Note: The value of acceleration due to gravity decreases both for a height and a depth from the surface of the earth. If the height is not negligible compared to the radius of the earth then we need to use the general formula g=g(1+hRe)2{g}'=g{{(1+\dfrac{h}{{{R}_{e}}})}^{-2}} for a depth of hh the body experiences gravitational attraction only due to the inner solid sphere, that is why the acceleration due to gravity also decreases with depth. For a height from the surface of the earth, the acceleration due to gravity decreases due to a larger distance.