Solved Problem on Kepler's Laws and Gravitation
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Calculate the acceleration due to gravity on a space station orbiting the Earth at a distance of 360 km from the surface (e.g. the International Space Station - ISS). Data: radius of Earth 6.37×103 km, mass of Earth 5.97×1024 kg, and Universal Gravitational Constant 6.67×10−11 N.m2/kg2.
Note: e.g. is the abbreviation of the Latin expression "exempli gratia" which means
"for example".
Problem data:
- Distance from the station to the surface of Earth: H = 360 km;
- Radius of Earth: R = 6.37×103 km;
- Mass of Earth: M = 5.97×1024 kg;
- Universal Gravitational Constant: G = 6.67×10−11 N.m2/kg2.
The size of the space station and the astronaut can be neglected relative to the distance they are from
the surface of Earth. On both bodies, the same gravitational force of attraction of the Earth acts,
\( {\vec{F}}_{G} \).
The distance from the station to the center of the Earth (Figure 1), is the sum of the radius of Earth,
R, and the distance from the station to the surface of Earth, H.
Solution
Applying Newton's Second Law to an astronaut of mass m in a circular orbit around the Earth
\[
\begin{gather}
\bbox[#99CCFF,10px]
{{\vec{F}}_{cp}=m{\vec{a}}_{cp}}
\end{gather}
\]
we have, in magnitude, that the only force acting on the astronaut (and on the space station) is the
gravitational attraction force of the Earth given by Newton's Law of Universal Gravitation
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{G}=G\frac{Mm}{r^{2}}}
\end{gather}
\]
equating the expressions above, we have the acceleration of a body orbiting the Earth
\[
\begin{gather}
G\frac{M\cancel{m}}{r^{2}}=\cancel{m}a_{cp}
\end{gather}
\]
canceling the mass m on both sides of the equality, the distance of the astronaut from the center of
the Earth is given by r=R+H
\[
\begin{gather}
a_{cp}=G\frac{M}{(R+H)^{2}}
\end{gather}
\]
substituting problem data
\[
\begin{gather}
a_{cp}=6.67\times 10^{-11}\times \frac{5.97\times 10^{24}}{\left(6.37\times 10^{6}+0.36\times 10^{6}\right)^{2}}\\[5pt]
a_{cp}=\frac{3.98\times 10^{14}}{\left(6.73\times 10^{6}\right)^{2}}\\[5pt]
a_{cp}=\frac{3.98\times 10^{14}}{4.53\times 10^{13}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{cp}=8.79\;\text{m/s}^{2}}
\end{gather}
\]
Notes:
1 – The variation of the acceleration on the space station relative to the acceleration on the Earth's surface is
Figure 2
Figure 3
1 – The variation of the acceleration on the space station relative to the acceleration on the Earth's surface is
\[
\begin{gather}
\frac{9.81-8.79}{9.81}=\frac{1.02}{9.81}=0.10=\frac{10}{100}=10\text{%}
\end{gather}
\]
The acceleration on the space station is only 10% less than what we feel on Earth.
2 – If the acceleration on the space station is not zero why do astronauts float?
The question can be understood based on a situation imagined by Isaac Newton (1642-1727) in his book A Treatise of the System of the World - 1728. Assume a cannon mounted on a very high mountain on Earth (Figure 2), if a shot is fired horizontally with low speed the bullet falls to Earth attracted by the force of gravity, if another shot is fired with more speed the bullet can go farther, but still falls to the Earth, when the cannon fires the shot with such a speed that the bullet goes around the Earth entering a circular orbit it will always be falling towards the Earth without ever arriving. This is the situation where the astronaut and the space station both fall toward Earth, this situation is comparable to a person in a free fall in an elevator (Figure 3).
The question can be understood based on a situation imagined by Isaac Newton (1642-1727) in his book A Treatise of the System of the World - 1728. Assume a cannon mounted on a very high mountain on Earth (Figure 2), if a shot is fired horizontally with low speed the bullet falls to Earth attracted by the force of gravity, if another shot is fired with more speed the bullet can go farther, but still falls to the Earth, when the cannon fires the shot with such a speed that the bullet goes around the Earth entering a circular orbit it will always be falling towards the Earth without ever arriving. This is the situation where the astronaut and the space station both fall toward Earth, this situation is comparable to a person in a free fall in an elevator (Figure 3).
This is the reason why in devices that simulate free fall in amusement parks, people must wear seat
belts, when the person and the chair fall in free fall, the person "takes off" from the seat by
falling out of the device.
Figure 3
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .