A satellite orbiting the moon very near the surface has a period of 110 min.
- Using this information, together with the radius of the moon being 1.74 x 106 m, to calculate the free fall acceleration on the moon’s surface. (i.e. don’t use the moon’s mass in your calculation).
- Now, based on your answer, estimate the mass of the moon.
Period of rotation = 110 min = 6600 s
Angular velocity, ω = (2π/6600) rad/sec
Centripetal force = Fc
Fc = m_s.v^2/r = m_s.ω^2.r (m_s = mass of satellite)
Gravitaional force = m_s.g_m (g_m = gravitational accln on moon)
Garvitational force = Centripetal force
m_s.g_m = m_s. ω^2.r
g_m = ω^2.r
g_m = (2π/6600)^2.(1.74*10^6)
g_m = 1.577 m^2/s
We can estimate the mass using the equation
Fg = G.(M1.M2).r^2, where
Fg = gravitational force of attraction between the two bodies, M1 and M2.
G = Gravitational constant = 6.67*10^(-11) N(m/kg)^2
M1 = mass of satellite = m_s
M2 = mass of moon = m_m
r = distance between the centres of mass of the two bodies = radius of moon = r_m
Using Fg = Fc = m_s*g_m,
m_s*g_m = G*(m_s*m_m)/(r_m)^2
g_m = G*(m_m)/(r_m)^2
m_m = g_m*(r_m)^2/G
m_m = 1.577*(1.74*10^6)^2/(6.67*10^(-11))
m_m = 1.577*(1.74)^2/(6.67)*10^(23)
m_m = 7.158*10^22 kg