Chapter 17, Problem 5P

Phobos obits Mars at a distance of 9376 km from the center of the planet and has a period of 0.3 189 days. Assume Phobos’s orbit is circular. Calculate the mass of Mars. (Hints: Use the circular orbit velocity formula in Reasoning with Numbers 4-1; make sure to convert relevant quantities to units of meters, kilograms. and seconds.)

To determine

To calculate:

The mass of Mars based on the distance and orbit angle of Phobos

Answer to Problem 5P

The mass of Mars = $6.45x{10}^{23}$ Kg.

Explanation of Solution

Given Information:

Radius of orbit = 9376 Km

Period of Orbit = 0.3189 Days

Formula Used:

Circular orbit velocity = $V_S=\sqrt{\frac{G.M}{r}}$

Also, V_S=distance/time

Calculation:

Assume Phobos (Moons) orbit to be circular.

First convert the quantities into proper units

  $1) Radiusoforbit=9376Km\begin{array}{l} =9376x1000m\hfill \\ =9.376x{10}^{6}m\hfill \end{array}$

  $2) Periodoforbit=0.3189Days\begin{array}{l}\begin{array}{l} \\ =0.3189x\left(24hr\right)x\left(60min\right)x\left(60sec\right)\end{array}\hfill \\ =27552.96Sec\hfill \end{array}$

Now,

Circular Velocity = $V_S=\sqrt{\frac{G.M}{r}}$ -------------------------------- (Assume moon is circular)

Where, G = Gravitational Constant = $6.67x{10}^{-11}{m}^3/s^{2}Kg$

r = Radius of Planet. (m)

m= Mass of Planet (Kg)

Rewriting in terms of m

∴ m = $\frac{Vs^2.r}{G}$

For finding circular velocity that is distance covered per unit time.

∴ V_S =Distance/time

Distance between but circumference of orbit.

∴$\begin{array}{l}circumferenceC=2\pi r\hfill \\ =2x\pi x9.376x{10}^6\hfill \\ {}^{ }=58911145.44\hfill \\ =5.9x{10}^{7}m\hfill \end{array}$

∴ Moon must cover 5.9 x 10⁷ m distance in 27552.96 sec

∴ its circular velocity must be,

  $V_S=\frac{5.9x10^7}{27552.9}=2141.33m/s$

∴ put all the Values in mass of Mars formula,

  $m=\frac{Vs^2.r}{G}$

= $\frac{2141.33X2141.33X9.376X10^6}{6.67X10^-11}$

= $6.45x{10}^{23}Kg$

Conclusion:

Mass of Mars is $6.45x{10}^{23}$ kg, which is around $10.7\%$ of Earth’s mass.