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 $1.8\times {10}^{7}m$and its rotation period to be role="math" localid="1648535932335" $22.3hours.$You have previously determined that the planet orbits $2.2\times \times {10}^{11}m$from its star with a period of $402earthdays$. Once on the surface you find that the free-fall acceleration is $12.2m/s^2.$What is the mass of (a) the planet and (b) the star?

Short answer

a. The mass of the planet is$1.48\times {10}^{25}kg.$

b. The mass of the star isrole="math" localid="1648535846026" $5.22\times {10}^{30}kg.$

Step-by-step solution

Given information a.

Radius of planet = $0.9\times {10}^{7}m$, time period =role="math" localid="1648536615774" $402days$, free-fall acceleration =$12.2m/s^2.$

Calculation a.

Using the formula of gravitational law we get :

$F=\frac{GmM}{r^2}=ma$

Where, F is gravitational force, M is mass of planet and r is the radius between masses.

Solving for acceleration we get :

$$\begin{gathered} a=\frac{GM}{r^2} \\ 12.2m/s^2=\frac{(6.67\times {10}^{-11}N{m}^2/k{g}^2)M}{{\left(0.9\times {10}^7\mathrm{m}\right)}^2} \\ M=1.48\times {10}^{25}\mathrm{kg} \end{gathered}$$

Final answer a.

The mass of the planet is$1.48\times {10}^{25}kg.$

Given information b.

Orbit distance = $2.2\times {10}^{11}m,$time period =$402days=3.473\times {10}^{7}s.$

Calculation b.

Using Kepler's third law of planetary motion, the time period is given by :

$T^2=\frac{4{\pi }^2{R}^3}{GM}$

Where, T is the time period, R is the radius of orbit of star, G is universal gravitational constant, M is mass of star.

$$\begin{gathered} T^2=\frac{4{\pi }^2{R}^3}{GM} \\ M=\frac{4{\pi }^2{R}^3}{G{T}^2} \\ M=\frac{4{\pi }^2{\left(2.2\times {10}^{11}m\right)}^3}{(6.67\times {10}^{-11}N{m}^2/k{g}^2){\left(34732800s\right)}^2} \\ M=5.22\times {10}^{30}\mathrm{kg} \end{gathered}$$

Final answer b.

The mass of the star is$5.22\times {10}^{30}kg.$