What would be the orbital speed and period of a satellite in orbit 1.50 ✕ 10^8 m above the Earth?
To determine the orbital speed and period of a satellite, we can use the equation:
v = √(GM/r)
Where:
v = orbital speed of the satellite
G = gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
M = mass of the Earth (approximately 5.972 × 10^24 kg)
r = radius from the center of the Earth to the satellite's orbit
First, we need to convert the given height of the satellite above the Earth's surface to the radius from the Earth's center. The radius (R) of the Earth is approximately 6.371 × 10^6 m, so the radius from the center to the satellite's orbit can be calculated as follows:
r = R + h
Where:
h = height above the Earth's surface
Plugging in the values:
r = 6.371 × 10^6 + 1.50 × 10^8 = 1.5631 × 10^8 m
Now we can calculate the orbital speed using the formula:
v = √(GM/r)
Plugging in the values:
v = √(6.67430 × 10^-11 × 5.972 × 10^24 / 1.5631 × 10^8)
Calculating this expression will give us the orbital speed of the satellite.
The orbital period (T) of a satellite is given by the equation:
T = 2πr/v
Plugging in the values:
T = 2π(1.5631 × 10^8) / v
Calculating this expression will give us the orbital period of the satellite.
By carrying out these calculations, you will be able to determine the orbital speed and period of the satellite in its given orbit.