A satellite is placed 39875 m above the Moon's SURFACE. (M=7.35e22 kg, R=1.74e6 m) Find the Period of the satellite in seconds. (assume a circular orbit)
r = .039875*10^6 +1.74*10^6 = 1.78*10^6
G = 6.67*10^-11
M = 7.35*10^22
F = m v^2/r
G M m /r^2 = m v^2/r
v^2 = G M /r
T^2 = (2pi)^2 * r^2/v^2
T^2 = (2pi)^2 * r^3 /GM
T^2 =
39.5*5.64*10^18/(6.67*10^-11*7.35*10^22)
T^2 = 4.54 * 10^7 = 45.4 * 10^6
T = 6.74 * 10^3 = 6740 seconds (about 1.8 hours)
thank you so much, now i understand
To find the period of the satellite's orbit, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period (T) of a satellite or planet is directly proportional to the cube of its average distance (r) from the center of the body being orbited.
Mathematically, this can be represented as:
T^2 = k * r^3
Where T is the period, r is the average distance, and k is the constant of proportionality.
In this case, the satellite is placed at an altitude above the Moon's surface. To find the average distance, we need to add the radius of the Moon to the altitude of the satellite. So, the average distance (r) is given by:
r = R + altitude
Given:
R (radius of the Moon) = 1.74e6 m
altitude = 39875 m
Let's calculate the average distance (r):
r = 1.74e6 + 39875
r = 1.78e6 m
Now, we can substitute the value of r into Kepler's Third Law equation:
T^2 = k * (1.78e6)^3
To find the constant k, we can rearrange the equation as follows:
k = T^2 / (1.78e6)^3
Given that the mass of the Moon (M) is 7.35e22 kg, we can use the gravitational constant (G) to find the constant k:
G = 6.67e-11 N(m/kg)^2
k = (G * M) / (4π^2)
Now, we can substitute the values of G, M, and r into the equation to find k:
k = (6.67e-11 * 7.35e22) / (4 * π^2)
k ≈ 43782.48 m^3/s^2
Now, we can solve for T by rearranging the equation:
T^2 = (43782.48) * (1.78e6)^3
T^2 ≈ 7.515e24
Taking the square root of both sides to solve for T:
T ≈ √(7.515e24)
T ≈ 2.742e12 seconds
Therefore, the period of the satellite in seconds is approximately 2.742e12 seconds.