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From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft traveling in a circular orbit 261 km above the earth's surface. Express W in both (a) newtons and (b) pounds.

Sagot :

Manetho

Answer:

[tex]W=\frac{773}{4.45}=173.76 l b f[/tex]

Explanation:

[tex]W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}[/tex]

The law of gravitation

[tex]G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)[/tex]

Universal gravitational constant [S.I. units]

[tex]m_{e}=5.976\left(10^{24}\right) k g[/tex]

Mass of Earth [S.I. units]

[tex]m=89 kg[/tex]

Mass of a man in a spacecraft [S.I. units]

[tex]R=6371 \mathrm{~km}[/tex]

Earth radius [km]

Distance between man and the earth's surface

[tex]h=261 \mathrm{~km} \quad[\mathrm{~km}][/tex]

ESULT [tex]W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}[/tex]

[tex]W=\frac{773}{4.45}=173.76 l b f[/tex]

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