To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:
[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]
Here,
[tex]v_{o} =[/tex] Orbital Velocity
[tex]\lambda_{max} =[/tex] Maximal Wavelength
[tex]\bar{\lambda}} =[/tex] Average Wavelength
c = Speed of light
Replacing with our values we have that,
[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]
Note that the average signal is 3.000000m
[tex]v_o = 36 km/s[/tex]
Now using the definition about centripetal acceleration we have,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Orbit Velocity
r = Radius of Orbit
Replacing with our values,
[tex]a = \frac{(36km/s)^2}{100000km}[/tex]
[tex]a= 0.01296km/s^2[/tex]
[tex]a = 12.96m/s^2[/tex]
Applying Newton's equation for acceleration due to gravity,
[tex]a =\frac{GM}{r^2}[/tex]
Here,
G = Universal gravitational constant
M = Mass of the planet
r = Orbit
The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,
[tex]M = \frac{ar^2}{G}[/tex]
[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]
[tex]M = 1.943028*10^{27}kg[/tex]
Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]
