Given: G=6.67259*10^ -11 Nm^ 2 /kg^ 2 The acceleration of gravity on the surface of a planet of radius R = 7040l is 11.1 m/s ^ 2 What is the period of a satellite in circu lar orbit h = 15488kn above the surface? Answer in units of
To find the period of a satellite in a circular orbit, we can use Kepler's third law of planetary motion, which states that the square of the period of an orbit is proportional to the cube of the semi-major axis of the orbit.
In this case, the semi-major axis can be obtained by adding the radius of the planet to the altitude of the satellite above the surface:
a = R + h
First, let's convert the given values to standard SI units:
R = 7040 * 1000 m = 7.04 * 10^6 m
h = 15488 * 1000 m = 1.5488 * 10^7 m
Now, we can calculate the semi-major axis:
a = R + h = 7.04 * 10^6 m + 1.5488 * 10^7 m = 2.25328 * 10^7 m
Next, we need to find the square of the period (T^2) using the formula:
T^2 = (4π^2 / GM) * a^3
where G is the gravitational constant G = 6.67259 * 10^-11 Nm^2/kg^2 and M is the mass of the planet. However, since we don't have the mass of the planet, we can use the acceleration of gravity on the surface as a substitute.
The acceleration of gravity at the surface (g) can be found using Newton's law of universal gravitation:
g = GM / R^2
Rearranging this equation, we have:
M = g * R^2 / G
Substituting the given values:
g = 11.1 m/s^2 (acceleration of gravity on the planet's surface)
G = 6.67259 * 10^-11 Nm^2/kg^2
M = (11.1 m/s^2) * (7.04 * 10^6 m)^2 / (6.67259 * 10^-11 Nm^2/kg^2)
M ≈ 6.0356 * 10^24 kg
Now, plug the values of G, M, and a into the equation to find T^2:
T^2 = (4π^2 / G * M) * a^3
T^2 = (4π^2 / (6.67259 * 10^-11 Nm^2/kg^2 * 6.0356 * 10^24 kg)) * (2.25328 * 10^7 m)^3
Calculating this expression gives:
T^2 ≈ 3.4478 * 10^8 s^2
Finally, to find the period (T), we take the square root:
T ≈ √(3.4478 * 10^8 s^2)
T ≈ 18555 s
In units of hours, this is:
T ≈ 5.15 hours
Therefore, the period of the satellite in a circular orbit 15,488 km above the surface of the planet is approximately 5.15 hours.