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 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 59.7 m/s2. What are the mass of (a) the planet and (b) the star?

Answer 1

Answer:

Part a)

$M = 7.25 \times 10^{25} kg$

Part b)

$M = 2\times 10^{30} kg$

Explanation:

Part a)

As we know that the diameter of the planet is given as

$d = 1.8 \times 10^7 m$

now the radius of the planet is given as

$r = 9 \times 10^6 m$

now we know that the acceleration due to gravity of the planet is given by the equation

$g = \frac{GM}{r^2}$

here we know that

$g = 59.7 m/s^2$

now from above equation we have

$59.7 = \frac{(6.67 \times 10^{-11})M}{(9\times 10^6)^2}$

now we have

$M = 7.25 \times 10^{25} kg$

Part b)

Now by kepler's law we know that

time period of revolution of planet about the star is given as

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

so we have

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

now we have

$\frac{1^2}{402^2} = \frac{(1.5 \times 10^11)^3}{r^3}}$

$rr = 8.17 \times 10^{12} m$

mula of time period

$402\times (365\times 24 \times 3600) = 2\pi \sqrt{\frac{(8.17\times 10^12)^3}{(6.67\times 10^{-11})M}}$

Now we have

$M = 2\times 10^{30} kg$