Respuesta :
Use Newton's Law of Universal Gravitation to find the gravitational force (which is the same as the weight) that a 68.0 kg person feels due to the interaction with Earth.
The force between two bodies with masses M and m separated by a distance r is:
[tex]F=G\frac{Mm}{r^2}[/tex]Where G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2}[/tex]Use M=5.97*10^24kg for the mass of the Earth, and R=6.371*10^6m for the radius of the Earth.
A)
For a person situated on the surface of the Earth, we can consider it to be situated at a distance equal to the radius of the Earth from all its mass. Then:
[tex]\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(5.97\times10^{24}\operatorname{kg})(68.0\operatorname{kg})}{(6.371\times10^6m)^2} \\ =667N \end{gathered}[/tex]B)
For a person situated 8,848 meters above the sea level, add 8,848m to the radius of the Earth in the formula:
[tex]\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(5.97\times10^{24}\operatorname{kg})(68.0\operatorname{kg})}{(6.371\times10^6m+8848m)^2} \\ =665N \end{gathered}[/tex]C)
For a person situated at 2.5 times the radius of the Earth from its center, multiply the radius of the earth by 2.5 in the formula:
[tex]\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(5.97\times10^{24}\operatorname{kg})(68.0\operatorname{kg})}{(2.5\times6.371\times10^6m)^2} \\ =107N \end{gathered}[/tex]