# You are the science officer on a visit to a distant solar system. Prior to landing on a planet,...

## Question:

You are the science officer on a visit to a distant solar system. Prior to landing on a planet, you measure its diameter to be {eq}1.90 \times 10^7 {/eq} m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits {eq}2.60 \times 10^{11} {/eq} m from its star with a period of 402 Earth days. Once on the surface, you find that the free-fall acceleration is 11.3 {eq}m/s^2. {/eq}

a) What is the mass of the planet?

b) What is the mass of the star?

## Acceleration Due to Gravity and Orbital Period of a Planet

Acceleration due to gravity of a planet is known, the mass of the planet can be calculated using the equation {eq}g = \dfrac { G M } { R^2 } {/eq}. Here *G* and *R* are the gravitational constant and radius of the planet. Orbital period *T* of the planet and radius *r* of the orbit of the planet around the star will help us to calculate the mass of the star with the equation {eq}T = \dfrac { 2\pi r } { v } {/eq}. Here {eq}v = \sqrt { \dfrac { G M_s } { r }} {/eq} is the orbital speed of the planet about the star of mass {eq}M_s {/eq}

## Answer and Explanation:

**Given points **

- Diameter of the planet {eq}d = 1.90 \times 10^7 \ m {/eq}

- Radius of the orbit of the planet about the distance star {eq}r = 2.60 \times 10^{11} \ m {/eq}

- Orbital period of the planet about the star {eq}T = 402 \ days \\ T = 402 \times 24 \times 3600 = 3.47 \times 10^7 \ s {/eq}

- Free fall acceleration on the surface of the planet {eq}g_p = 11.3 \ m/s^2 {/eq}

- Gravitational constant {eq}G = 6.674 \times 10^{-11} \ m^3/ kg . s^2 {/eq}

**Part a )**

Let {eq}M_p {/eq} be the mass of the planet.

Radius of the planet {eq}R = \dfrac { d } { 2 } = 9.50 \times 10^6 \ m {/eq}

Then we can express free fall acceleration of the surface of the planet as {eq}g_p = \dfrac { G M_p } { R^2 } {/eq}

So mass of the planet {eq}M_p = \dfrac { g_p R^2 } { G } \\ M_p = \dfrac { 11.3 \times (9.50 \times 10^6 )^2 } { 6.674 \times 10^{-11} } \\ M_p = 1.53 \times 10^{25} \ kg {/eq}

**Part b)**

Let {eq}M_s {/eq} be the mass of the star.

Then orbital speed of the planet about the star can be expressed as {eq}v = \sqrt { \dfrac { G M_s } { r } } {/eq}

Then orbital period of the planet {eq}T = \dfrac { 2 \pi r } { v } {/eq}

Squaring the expression and substituting the value of the square of the orbital speed in the above expression

We get {eq}T^2 = \dfrac { 4 \pi^2 r^2 } { ( \dfrac { G M_s } { r } ) } {/eq}

So mass of the star {eq}M_s = \dfrac { 4 \pi^2 r^3 } { G T^2 } \\ M_s = \dfrac { 4 \times \pi^2 \times ( 2.60 \times 10^{11} )^3 } { 6.674 \times 10^{-11} \times ( 3.47 \times 10^7)^2 } \\ M_s = 8.63 \times 10^{30} \ kg {/eq}