Question

identify how the period of a planet varies with each factor below.
distance, r, of the planet from the sun
the sun’s mass, m_s

Explanation:

Step1: Recall Kepler's Third Law

Kepler's third law (modified by Newton) for a planet orbiting the Sun is given by \( T^{2}=\frac{4\pi^{2}}{Gm_{s}}r^{3} \), where \( T \) is the period, \( r \) is the orbital radius, \( m_{s} \) is the Sun's mass, and \( G \) is the gravitational constant.

Step2: Analyze effect of \( r \) on \( T \)

From the formula \( T^{2}\propto r^{3} \) (since \( \frac{4\pi^{2}}{Gm_{s}} \) is constant for a given Sun), we can rewrite it as \( T\propto r^{\frac{3}{2}} \). So if \( r \) increases, \( T \) increases as \( r^{\frac{3}{2}} \), and if \( r \) decreases, \( T \) decreases as \( r^{\frac{3}{2}} \).

Step3: Analyze effect of \( m_{s} \) on \( T \)

From \( T^{2}=\frac{4\pi^{2}}{Gm_{s}}r^{3} \), we can solve for \( T \): \( T = \sqrt{\frac{4\pi^{2}r^{3}}{Gm_{s}}}\propto\frac{1}{\sqrt{m_{s}}} \) (since \( r \) and other constants are fixed for the relationship with \( m_{s} \)). So if \( m_{s} \) increases, \( T \) decreases, and if \( m_{s} \) decreases, \( T \) increases.

Answer:

For distance \( r \): Period \( T \) varies as \( T \propto r^{\frac{3}{2}} \) (increases with \( r^{\frac{3}{2}} \) when \( r \) increases, decreases with \( r^{\frac{3}{2}} \) when \( r \) decreases).
For Sun's mass \( m_{s} \): Period \( T \) varies as \( T \propto \frac{1}{\sqrt{m_{s}}} \) (decreases when \( m_{s} \) increases, increases when \( m_{s} \) decreases).