What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 208 km above Earth's surface? (b) What is the period of revolution?
I got a to be 7786 m/s^2 but im not sure how to get b.
m/s is speed
m/s^2 is acceleration
G m Me/r^2 = m v^2/r
G Me/r = v^2
r = .208*10^6 + 6.38*10^6 = 6.59 * 10^6 meters
v^2 = 6.67*10^-11*5.98*10^24/6.59*10^6
v^2 = 6.05*10^7 = 60.5*10^6
v = 7780 m/s agree
now
circumference = 2 pi r
time = circumference/speed
is the time is seconds or minutes?
To find the linear speed required for a satellite in a circular orbit at a given altitude, you can use the following formula:
v = √(GM/r)
where:
- v is the linear speed of the satellite,
- G is the gravitational constant (approximately equal to 6.674 × 10^(-11) m^3/(kg s^2)),
- M is the mass of the Earth (approximately equal to 5.972 × 10^24 kg),
- r is the radius of the Earth plus the altitude of the satellite (in this case, 208 km).
Now let's calculate the linear speed (v):
Step 1: Convert the altitude of the satellite to the same unit as the radius of the Earth (meters).
208 km = 208,000 m
Step 2: Calculate the radius of the Earth plus the altitude.
radius = radius of the Earth + altitude
radius = 6,371,000 m + 208,000 m
radius = 6,579,000 m
Step 3: Substitute the values into the formula to find the linear speed.
v = √((6.674 × 10^(-11) * 5.972 × 10^24) / 6,579,000)
After performing the calculation, v ≈ 7,682 m/s (rounded to the nearest whole number).
Now let's move on to finding the period of revolution (T), which represents the time it takes for the satellite to complete one orbit.
The formula for the period is given by:
T = (2πr) / v
where:
- T is the period of revolution,
- r is the radius of the Earth plus the altitude (6,579,000 m),
- v is the linear speed (7,682 m/s).
Step 1: Substitute the values into the formula to find the period.
T = (2π * 6,579,000) / 7,682
After performing the calculation, T ≈ 5,411 seconds (rounded to the nearest whole number).
Therefore, the linear speed for an Earth satellite in a circular orbit at an altitude of 208 km above the Earth's surface is approximately 7,682 m/s, and the period of revolution is approximately 5,411 seconds.