What must be the orbital speed of a satellite in a circular orbit 740 km above the surface of the earth?
The distance from the center of the earth will be
R = 740 + Re
= 6380 = 7120 km = 7.12*10^6 m
GMm/R^2 = mV^2/R
V = sqrt(GM/R)
G is the universal gravity constant and M is the mass of the earth. Solve for V
To find the orbital speed of a satellite, we can use the formula for circular motion: v = √(G * M / r), where v is the orbital speed, G is the gravitational constant, M is the mass of the Earth, and r is the radius of the satellite's orbit.
Step 1: Find the radius of the satellite's orbit
The altitude of the satellite above the Earth's surface is 740 km. To get the radius of the orbit, we need to add the radius of the Earth to the altitude of the satellite.
Radius of Earth = 6,371 km (approximately)
Radius of the satellite's orbit = Radius of Earth + Altitude of satellite
= 6,371 km + 740 km
= 7,111 km
Step 2: Calculate the orbital speed
Now that we have the radius of the satellite's orbit, we can calculate the orbital speed using the formula:
v = √(G * M / r)
Gravitational constant (G) = 6.67430 × 10^-11 m^3 kg^-1 s^-2 (standard value)
Mass of Earth (M) = 5.972 × 10^24 kg (standard value)
Radius of satellite's orbit (r) = 7,111 km = 7,111,000 meters (converting to meters)
v = √(6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg / 7,111,000 meters)
Simplifying the equation and calculating the value will give us the orbital speed of the satellite.