What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a period of exactly 24.0 hours (86,400 seconds). (Measurement is from the center of Earth.)
G = 6.67 × 10-11 N.m2/kg2
ME = 5.98 × 1024 kg
1 hr = 3,600 s
The period is
T=2•π• sqrt (R³ /G•M)
Solve for R
What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a period of exactly 24.0 hours (86,400 seconds). (Measurement is from the center of Earth.)
To determine the radius of a communications satellite's orbital path, we can use Kepler's third law of planetary motion, which states that the square of the period of an object's orbit is proportional to the cube of its average distance from the center of mass it orbits.
First, we need to find the average distance or radius of the orbit using the given period of 24.0 hours or 86,400 seconds.
Let's denote the radius as r.
Let's denote the period as T.
According to Kepler's third law, T^2 = (4π^2/GM)r^3, where T is the period, G is the gravitational constant, M is the mass of the central body (Earth), and r is the radius.
Given:
G = 6.67 × 10^-11 N.m^2/kg^2
M = 5.98 × 10^24 kg
T = 86,400 seconds
Rearranging the equation to solve for r, we get:
r = ((GMT^2)/(4π^2))^1/3
Substituting the given values:
r = ((6.67 × 10^-11 N.m^2/kg^2) * (5.98 × 10^24 kg) * (86400 s)^2) / (4π^2))^1/3
Simplifying the equation, we get:
r = 4.22 × 10^7 meters
Therefore, the radius of the satellite's orbital path is approximately 4.22 × 10^7 meters from the center of the Earth.