(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 129 km above Earth's surface? (b) What is the period of revolution?
All right, I swear I have done everything and cannot figure this out!
The velocity required to maintain a circular orbit around the Earth may be computed from the following:
Vc = sqrt(µ/r)
where Vc is the circular orbital velocity in feet per second, µ (pronounced meuw as opposed to meow) is the gravitational constant of the earth, ~1.407974x10^16 ft.^3/sec.^2, and r is the distance from the center of the earth to the altitude in question in feet. Using 3963 miles for the radius of the earth, the orbital velocity required for a 250 miles high circular orbit would be Vc = 1.407974x10^16/[(3963+250)x5280] = 1.407974x10^16/22,244,640 = 25,155 fps. (17,147 mph.) Since velocity is inversely proportional to r, the higher you go, the smaller the required orbital velocity.
The length of time it takes for a satellite to orbit the earth, its orbital period, varies with the altitude of the satellite above the earth's surface. The lower the altitude, the shorter the period. The higher the altitude, the longer the period. For example, the orbital period for a 100 mile high satellite is ~88 minutes; 500 miles ~101 minutes; 1000 miles ~118 minutes; 10,000 miles 9hr-18min; 22,238 miles 23hr-56min-4.09sec. A satellite in an equatorial orbit of 22,238 miles altitude remains stationary over a point on the Earth's equator and the orbit is called a geostationary orbit. A satellite at the same 22,238 miles altitude, but with its orbit inclined to the equator, has the same orbital period and is referred to as a geosynchronous orbit as it is in sync with the earth's rotation.
Not surprisingly, the velocity of a satellite reduces as the altitude increases. The velocities at the same altitudes described above are 25,616 fps. (17,426 mph) for 100 miles, 24,441 fps. (16,660 mph.) for 500 miles, 23,177 fps. (15,800 mph.) for 1000 miles, 13,818 fps. (9419 mph) for 10,000 miles, and 10,088 fps. (6877 mph.) for 22,238 miles.
Depending on your math knowledge, you can calculate the orbital period from a simple expression. You might like to try it out if you have a calculator.
The time it takes a satellite to orbit the earth, its orbital period, can be calculated from
T = 2(Pi)sqrt[a^3/µ]
where T is the orbital period in seconds, Pi = 3.1416, a = the semi-major axis of an elliptical orbit = (rp+ra)/2 where rp = the perigee (closest) radius and ra = the apogee (farthest) radius from the center of the earth, µ = the earth's gravitational constant = 1.407974x10^16 ft.^3/sec.^2. In the case of a circular orbit, a = r, the radius of the orbit. Thus, for a 250 miles high circular orbit, a = r = (3963 + 250)5280 ft. and T = 2(3.1416)sqrt[[[(3963+250)5280]^3]/1.407974x10^16] = ~5555 seconds = ~92.6 minutes.
This should enable you to solve your problem.
No worries! I'm here to help. Let's break down the problem and find the answers step by step.
(a) To find the linear speed of an Earth satellite in a circular orbit at a given altitude, we can use the formula for the orbital speed:
v = √(G * M / r),
where v is the linear speed, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the Earth (approximately 5.972 × 10^24 kg), and r is the distance from the satellite to the center of the Earth (which is the sum of the Earth's radius and the altitude of the satellite).
Given that the altitude is 129 km, the radius can be calculated as:
r = radius of the Earth + altitude
= 6,371 km + 129 km
= 6,500 km.
Now we can plug in these values into the formula to find the linear speed:
v = √(6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg / 6,500,000 m)
≈ 7,680 m/s.
Therefore, the linear speed of the satellite must be approximately 7,680 m/s.
(b) To find the period of revolution (T), we can use the formula:
T = 2πr / v,
where T is the period, r is the distance from the satellite to the center of the Earth (which we calculated as 6,500 km or 6,500,000 m), and v is the linear speed (which we calculated as 7,680 m/s).
Plugging in these values, we can calculate the period as follows:
T = 2π * 6,500,000 m / 7,680 m/s
≈ 5,393 seconds.
Therefore, the period of revolution is approximately 5,393 seconds.
I hope this helps! Let me know if you have any further questions.