A 581 kg satellite is in a circular orbit about the Earth at a height above the Earth equal to the Earth's mean radius. the orbital speed is 5590.98 m/s. Find the period of its revolution.
nevermind. i just figured it out.
To find the period of revolution of a satellite in circular orbit around the Earth, we can use the relation between the orbital speed, radius of the orbit, and the period.
The orbital speed of the satellite is given as 5590.98 m/s.
The radius of the orbit is equal to the height of the satellite above the Earth's surface, which is the Earth's mean radius. Let's denote the radius of the Earth as R. Therefore, the radius of the orbit is 2R (since the satellite is at a height equal to the Earth's mean radius).
Let's denote the period of revolution as T.
Now, let's use the formula for the centripetal force acting on the satellite:
F = (m * v^2) / r,
where F is the gravitational force, m is the mass of the satellite, v is the orbital speed, and r is the radius of the orbit.
The gravitational force is given by:
F = (G * M * m) / r^2,
where G is the gravitational constant (approximately 6.67430 x 10^-11 Nm^2/kg^2) and M is the mass of the Earth (approximately 5.972 x 10^24 kg).
Since the gravitational force provides the centripetal force for the satellite's motion, we can equate the two equations:
(m * v^2) / r = (G * M * m) / r^2.
Now, we can solve for the period T using the relation between speed, distance, and time:
v = (2 * π * r) / T.
Let's substitute the value of r from above into this equation:
5590.98 = (2 * π * (2R)) / T.
Now, we can solve for T:
T = (2 * π * (2R)) / 5590.98.
Substituting the values of R (mean radius of the Earth, approximately 6371 km), π (pi, approximately 3.14159), and solving the equation, we can calculate the period T in seconds.