Question

Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth.

Answer 1

Therefore, the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth is approximately 3.07 x 10^3 m/s.

Step-by-step explanation:

A geosynchronous circular orbit is an orbit in which a satellite revolves around the Earth once every 24 hours so that it appears to be stationary in the sky relative to an observer on the ground. The radius of such an orbit is known as the geostationary radius and is approximately 42,164 km or 3.58 x 10^7 m above the surface of the Earth.

To find the orbital speed of a satellite in a geosynchronous circular orbit, we can use the formula:

v = (GM / r)^0.5

Where v is the orbital speed, G is the gravitational constant (6.674 x 10^-11 N m^2/kg^2), M is the mass of the Earth (5.97 x 10^24 kg), and r is the distance from the center of the Earth to the satellite (in this case, r = 3.58 x 10^7 m + radius of the Earth).

The radius of the Earth is approximately 6,371 km or 6.371 x 10^6 m. Therefore, the distance from the center of the Earth to the satellite is:

r = 3.58 x 10^7 m + 6.371 x 10^6 m

r = 4.217 x 10^7 m

Now we can plug in the values for G, M, and r into the formula and solve for v:

v = (GM / r)^0.5

v = [(6.674 x 10^-11 N m^2/kg^2) x (5.97 x 10^24 kg) / (4.217 x 10^7 m)]^0.5

v = 3.07 x 10^3 m/s

Therefore, the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth is approximately 3.07 x 10^3 m/s.

Answer 2

Step-by-step explanation:

The orbital speed of a satellite in a circular orbit around the Earth can be found using the following formula:

v = sqrt(G*M/R)

where G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth to the satellite's orbit.

For a geosynchronous orbit, the satellite has a period of 24 hours, which means it completes one orbit in 24 hours. This corresponds to an orbital radius of:

R = 3.58 x 10^7 m + 6.38 x 10^6 m = 4.22 x 10^7 m

where 6.38 x 10^6 m is the radius of the Earth.

The mass of the Earth is approximately 5.97 x 10^24 kg, and the gravitational constant is approximately 6.6743 x 10^-11 N*(m/kg)^2. Substituting these values into the equation above, we get:

v = sqrt(GM/R) = sqrt(6.6743 x 10^-11 N(m/kg)^2 * 5.97 x 10^24 kg / 4.22 x 10^7 m) = 3074 m/s

Therefore, the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 10^7 m above the surface of the Earth is approximately 3074 m/s.