Solved Problem on Kepler's Laws and Gravitation
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Mars is 52% further from the Sun than Earth. Calculate, in Earth years, the period of the revolution of Mars around the Sun.


Problem data:
  • Distance from Earth to Sun:    dEarth = R;
  • Period of revolution of Earth:    T Earth = 1 year.
Problem diagram:

Mars is 52%=0.52 further from the Sun than the Earth, 0.52R, the distance from Mars to the Sun is the distance from Earth to the Sun plus the extra distance, dMars = R+0.52R = 1.52R.

Figure 1

Solution

Using Kepler's Third Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{T_{Earth}^{2}}{R_{Earth}^{3}}=\frac{T_{Mars}^{2}}{R_{Mars}^{3}}} \end{gather} \]
solving for the period of revolution of Mars around the Sun, which is the unknown
\[ \begin{gather} T_{Mars}^{2}=\frac{T_{Earth}^{2}}{R_{Earth}^{3}}R_{Mars} \end{gather} \]
substituting the data
\[ \begin{gather} T_{Mars}^{2}=\frac{(1)^{2}\times (1.52R)^{3}}{R^{3}}\\[5pt] T_{Mars}^{2}=\frac{3.51\cancel{R^{3}}}{\cancel{R^{3}}} \end{gather} \]
canceling R3 into the numerator and denominator
\[ \begin{gather} T_{Mars}=\sqrt{3.51\;} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {T_{Mars}=1.87\;\text{earth years}} \end{gather} \]
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