A spy satellite is in circular orbit around Earth. It makes one revolution in 6.01 hours.
how high above earth?
acceleration?
change time to seconds for the period.
v=2PI h/period
v^2=(2PI)^2/period^2
v^2/r= 9.8*(re^2/r^2) where r is the orbital radius.
solve for r, then altitude is r-re.
To find out the height above Earth and the acceleration of the spy satellite, we'll need to use the following formulas:
1. For the height above Earth:
h = R + R0
where h is the height above Earth, R is the radius of the Earth, and R0 is the radius of the circular orbit.
2. For the acceleration:
a = (4π²R) / T²
where a is the acceleration, R is the radius of the Earth, and T is the period (time for one revolution) of the satellite.
Let's calculate these step by step:
Step 1: Find the radius of the Earth
The average radius of the Earth is approximately 6,371 kilometers (or 6,371,000 meters).
Step 2: Find the radius of the orbit (R0)
The radius of the orbit is equal to the height above Earth. Substituting it into the first formula, we have:
h = R + R0
R0 = h - R
Step 3: Calculate the height above Earth (h)
Since we don't have the height above Earth information in the given question, we can't calculate the exact value using the given information.
Step 4: Calculate the acceleration (a)
Using the given period of the satellite (T = 6.01 hours) and substituting it into the formula, we have:
a = (4π²R) / T²
Now, I can calculate the acceleration if you provide me with the radius of the Earth and the height above Earth for the circular orbit.
To determine the height above Earth's surface at which a satellite is orbiting, we can use the following equation:
r = R + h
Where:
r is the distance from the center of the Earth to the satellite's position
R is the radius of the Earth (approximately 6,371 kilometers)
h is the altitude or height above the Earth's surface
Since the satellite is in a circular orbit, its distance from the center of the Earth will be constant. Therefore, the distance travelled by the satellite during one revolution will be equal to the circumference of its circular orbit.
C = 2πr
Given that the satellite completes one revolution in 6.01 hours, we can determine the circumference using the equation:
C = v × t
Where:
v is the orbital velocity of the satellite
t is the time taken for one revolution
Rearranging the equation, we have:
v = C / t
Substituting the known values, we have:
v = 2πr / t
Now, let's calculate the radius of the circular orbit and subsequently the height above Earth's surface using the given information.
Dividing the circumference by the time of one revolution:
v = (2πr) / (6.01 hours)
To calculate the acceleration, we can use Newton's second law of motion. For an object in circular motion, the centripetal force is given by:
F = m × a
Where:
F is the gravitational force between the satellite and the Earth
m is the mass of the satellite (which is canceled out in the equation)
a is the centripetal acceleration required to keep the satellite in its circular orbit
The gravitational force between the satellite and the Earth can be expressed by the equation:
F = G × (m × M) / r^2
Where:
G is the gravitational constant (approximately 6.674 × 10^-11 N·m^2/kg^2)
M is the mass of the Earth (approximately 5.972 × 10^24 kg)
Setting the centripetal force equal to the gravitational force:
(G × (m × M) / r^2) = m × a
Canceling the mass of the satellite:
(G × M) / r^2 = a
Substituting the known values, we have:
(G × M) / r^2 = a
Now, let's calculate the acceleration using the gravitational force equation and the given values.