COMPLETELY STUCK? The period of rotation of the Moon is about 27 days, 8 hrs. The mass of the Moon is 7.35 x 1022 kg, and the radius of the Moon is 1.74 x 106 m. I wish to place a synchronous (period of orbit = period of the Moon’s rotation) satellite in orbit about the Moon.
a) Find the radius of the satellite’s orbit.
b) Find the height of the satellite above the Moon’s surface
a) the gravitational constant
G =6.67•10⁻¹¹ N•m²/kg²,
r=1.74•10⁶ m
M=7.35•10²² kg
ma=F
F =G•m•M/R²
ma=mv²/R
G•m•M/R² =mv²/R
v=sqrt(GM/R)
T=2πR/v = 2π•sqrt{R³/GM}
T²= 4π²•R³/GM
Solve for R.
b) h=R-r = …
To find the radius of the satellite's orbit around the Moon, you first need to find the orbital period of the Moon.
The period of the Moon's rotation is given as 27 days, 8 hours. However, it's important to convert this into seconds to ensure consistent units of time.
In a day, there are 24 hours and in an hour, there are 60 minutes. Therefore, the total number of seconds in the Moon's rotation period is:
27 days * 24 hours * 60 minutes * 60 seconds = 2,332,800 seconds
Now that we have the orbital period of the Moon, we can use Kepler's third law to find the radius of the satellite's orbit. Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis.
Let's denote the radius of the satellite's orbit around the Moon as R. Since we want the period of the satellite's orbit to be the same as the Moon's rotation period, we can set these two periods equal:
T_satellite = T_Moon
By substituting the values we have, we can solve for R:
R^3 / (R+1.74 x 10^6)^2 = 2,332,800^2
Simplifying this equation will give us the radius of the satellite's orbit around the Moon.
To find the height of the satellite above the Moon's surface, you can subtract the radius of the Moon from the radius of the satellite's orbit:
Height = R - 1.74 x 10^6
By plugging in the value of R obtained previously, you can calculate the height of the satellite above the Moon's surface.