A satalites escape velocity is 6.5 mi/sec, the radius of the earth is 3960 miles and the earth's gravitational constant(g) is 32.2 ft/sec^2 How far is the satelite from the surface of the earth?
To find the distance of the satellite from the surface of the earth, we need to calculate the height above the Earth's surface at which the satellite orbits. We can use the concept of escape velocity to find this distance.
The formula for escape velocity (Ve) is:
Ve = sqrt(2 * g * r)
Where:
Ve is the escape velocity,
g is the gravitational constant,
r is the radius of the Earth.
Given that the escape velocity (Ve) is 6.5 mi/sec, the radius of the Earth (r) is 3960 miles, and the gravitational constant (g) is 32.2 ft/sec^2, we can proceed to calculate the height of the satellite above the Earth's surface.
First, we need to convert the units to be consistent. Let's convert the radius to feet:
r = 3960 miles * 5280 ft/mile
r = 20,928,000 ft
Now, we can substitute the values into the formula:
6.5 mi/sec = sqrt(2 * 32.2 ft/sec^2 * 20,928,000 ft)
To solve for the distance, we need to square both sides of the equation:
(6.5 mi/sec)^2 = 2 * 32.2 ft/sec^2 * 20,928,000 ft
42.25 mi^2/sec^2 = 2 * 32.2 ft/sec^2 * 20,928,000 ft
Now, let's solve for the height (h) above the Earth's surface:
h = [(6.5 mi/sec)^2 / (2 * 32.2 ft/sec^2)] * 20,928,000 ft
Performing the calculation:
h ≈ 101,679,439 ft
Therefore, the satellite is approximately 101,679,439 feet above the surface of the Earth.