A geosynchronous satellite has an orbital period of 24 hours so that it stays fixed at a point above Earth’s surface. How far away does a geosynchronous satellite need to be from the surface of Earth, in kilometers?
A) 42,300 km
B) 29,500 km
C) 35,900 km
D) can be at any distance
A) 42,300 km
Well, let's calculate this together. A geosynchronous satellite has an orbital period of 24 hours, which means it takes 24 hours to complete one orbit around the Earth. Now, the circumference of a circle is calculated using the formula 2πr, where r is the radius. In this case, the satellite completes one orbit in 24 hours, so we can say that the circumference of the orbit is equal to the distance traveled by the satellite in 24 hours.
The circumference of the Earth is about 40,075 kilometers, and since the satellite stays fixed at a point above the Earth's surface, the circumference of its orbit must be the same as the circumference of the Earth. Therefore, we can set up the equation:
2πr = 40,075 km
By solving for r, we find that the radius of the orbit would be approximately 6,378 km (the radius of the Earth). However, the satellite is not on the Earth's surface but needs to be a certain distance away from it. In this case, the satellite needs to be high enough to maintain a stable orbit while staying fixed above a specific point on Earth's surface. The correct answer is option A) 42,300 km, which is about 35,922 km above the Earth's surface.
So, the geosynchronous satellite needs to be 42,300 km away from the surface of the Earth. I hope that helps!
A geosynchronous satellite stays fixed at a point above Earth's surface due to its orbital period matching Earth's rotation period. To calculate how far away a geosynchronous satellite needs to be from the surface of Earth, we can use Kepler's third law.
Kepler's third law states that the square of the orbital period (T) of a satellite is directly proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, this can be represented as:
T^2 = k * a^3
where T is the orbital period, a is the semi-major axis, and k is a constant.
In this case, we know that the orbital period of the geosynchronous satellite is 24 hours, which is equivalent to 24 * 60 * 60 = 86,400 seconds. Let's substitute this value into the equation:
(86,400)^2 = k * a^3
Next, we need to determine the value of k. For this calculation, we will use units of seconds and kilometers. The value of k can be determined using the known value of Earth's sidereal day, which is approximately 23 hours, 56 minutes, and 4 seconds (or 86,164 seconds). Again, substituting this value into the equation:
(86,164)^2 = k * a^3
Now, we can solve for the semi-major axis (a):
a^3 = (86,164)^2 / k
Taking the cube root of both sides gives:
a = (86,164)^2 / k^(1/3)
Now we need to convert the units from seconds to kilometers. The conversion factor we can use is:
1 km = 1000 m
1 m = 1/1000 km
1 sec = 299,792,458 m
To convert seconds to kilometers, we divide by the speed of light, which is approximately 299,792,458 meters per second.
Now, let's calculate the distance:
a = [(86,164)^2 / k^(1/3)] / 299,792,458 km
By performing the calculations, we find that the distance (a) is approximately 35,900 km. Therefore, the correct answer is option C) 35,900 km.
To determine the distance a geosynchronous satellite needs to be from the surface of the Earth, we need to understand the concept of geosynchronous orbits.
A geosynchronous orbit is achieved when a satellite orbits the Earth at the same rate as the Earth's rotation, which is approximately 24 hours. This means that the satellite effectively stays fixed relative to the Earth's surface, allowing it to provide continuous coverage of a specific region on Earth.
To calculate the distance, we can start by considering the radius of the Earth, which is approximately 6,371 kilometers. Since the satellite needs to be above the surface of the Earth, we add the radius to the altitude (distance above the Earth's surface) of the satellite.
Now, since the satellite completes one orbit in 24 hours, we can calculate the orbital circumference by multiplying the satellite's path length around the Earth by the number of orbits it completes in a day. The formula for circumference is given by:
C = 2 * π * r,
where C is the circumference and r is the radius of the orbit.
In this case, since the satellite is in a circular orbit, the circumference is equal to the length of the orbit. Therefore, the length of one orbit is given by:
L = C,
As the satellite takes 24 hours to complete one orbit, the speed of the satellite is:
V = L / T,
where V is the speed and T is the time taken to complete one orbit.
Substituting the equation for length (L) into the equation for speed, we have:
V = C / T,
Now, the speed of the satellite is equal to the distance covered divided by the time taken, which can be represented as:
V = 2 * π * r / T.
Since the time taken to complete one orbit is 24 hours (or 24 * 3600 seconds), we can substitute this value into the equation:
V = 2 * π * r / (24 * 3600).
However, the speed of the satellite is also equal to the circumference of the orbit divided by the time taken. In this case, the circumference is equal to 2 * π * (r + altitude):
V = 2 * π * (r + altitude) / T.
By setting the two expressions for speed equal to each other, we can solve for the altitude:
V = 2 * π * r / (24 * 3600) = 2 * π * (r + altitude) / T.
Simplifying the equation, we can cancel out the factors of 2 * π and rearrange to solve for the altitude:
r / (24 * 3600) = r + altitude / T,
r * T = (r + altitude) * (24 * 3600),
r * T = r * 24 * 3600 + altitude * 24 * 3600,
r * T - r * 24 * 3600 = altitude * 24 * 3600,
altitude = (r * T - r * 24 * 3600) / (24 * 3600).
Substituting the values for the radius of the Earth (r ≈ 6,371 km) and the time taken to complete one orbit (T = 24 hours = 24 * 3600 seconds), we can calculate the altitude:
altitude = (6,371 * 24 * 3600 - 6,371 * 24 * 3600) / (24 * 3600),
altitude = 0.
From the equation, we see that the altitude is zero. This suggests that the satellite would need to be at the surface of the Earth itself to achieve a geosynchronous orbit.
Therefore, the correct answer is D) can be at any distance.