What is the magnitude of the gravitational force between the earth and 1 kg object on its surface mass of the Earth is 6 into 10/24 kg and radius of the Earth is 6.4 into 10 6?

Text Solution

Solution : Given that <br> Mass of the body, m=1kg <br> Mass of the Earth, `M=6xx10^(24)kg`. <br> Radius of the earth, `R=6.4xx10^(6)` <br> Now and of the gravitational force between the earth and the body can be the <br> `F=G(Mxm)/(r^(2))=(6.67xx10xx6xx10xx1)/((6.4xx10^(6))^(2))` <br> `=(6.67xx6xx10)/(6.4xx64.4)=9.8N`.

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m

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From the universal law of gravitation, the magnitude of the gravitational force (F) between the earth and the body is given by,

$F=G\frac{M\times m}{R^2}$

where $G=6.67\times {10}^{-11}N{m}^2/k{g}^2$ is the universal gravitational constant.

Hence, magnitude of the gravitational force (F) between the earth and the body is given by,

$F=G\frac{M\times m}{R^2}=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 1}{(6.4\times {10}^6{)}^2}$ = 9.77 N = 9.8 N (approx).