A spy satellite is in circular orbit around Earth. It makes one revolution in 6.04 hours.
(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?
YOu start with centripetal acceleration= gravitational attraction.
w^2 * r= 9.8 (re^2/(r^2)) where re is the radius of the Earth, and r is the distance from the center of the Earth to the satellite.
w= 2PIr/period work in m/s, m
so W = (2PI(6.37x10^6))/21,744sec
6.04hr = 21,744 sec
and then for r = 117,367
I don't get what to do afterwards
To determine the height above Earth's surface and the acceleration of the spy satellite, we can make use of the following formulas:
(a) Height above Earth's surface:
The height above Earth's surface can be calculated using the equation for the radius of a circular orbit:
r = R + h
Where:
r = radius of the orbit
R = radius of Earth
h = height above Earth's surface
We know the period (T) of the orbit, which is the time it takes for one revolution. The formula relating the period to the radius is:
T = 2π √(r^3 / GMe)
Where:
G = the gravitational constant
Me = mass of Earth
Solving this equation for r, we can then calculate the height by subtracting the radius of Earth.
(b) Acceleration:
The acceleration experienced by an object in circular motion can be calculated using the equation:
a = v^2 / r
Where:
a = acceleration
v = velocity of the satellite
r = radius of the orbit
Let's calculate them step by step:
(a) Height above Earth's surface:
1. Convert the given period from hours to seconds.
T = 6.04 hours × 3600 seconds/hour = 21744 seconds
2. Calculate the radius of the orbit by rearranging the formula for period:
r = (GMe * T^2) / (4π^2)^(1/3)
3. Subtract the radius of Earth to determine the height above the surface.
(b) Acceleration:
1. Calculate the velocity of the satellite using the formula:
v = (2πr) / T
2. Substitute the values into the equation for acceleration.
Now let's compute the answers.