A satellite is launched into orbit 200 kilometers above the Earth.The orbital velocity of a satellite is given by the formula v=√GmE/r, where v is the velocity in meters per second, G is a given constant, mE is the mass of Earth, and r is the radius of the satellite"s orbit. The radius of Earth is 6,380,000 meters. What is the radius of the satellite's orbit in meters?
To find the radius of the satellite's orbit, we need to rearrange the given formula and solve for "r".
The given formula is: v = √(GmE/r)
We need to rearrange it to solve for "r". To do this, we square both sides of the equation to eliminate the square root:
v^2 = GmE/r
Now, we can rearrange the equation to solve for "r":
r = GmE / v^2
Given that the satellite is orbiting 200 kilometers (or 200,000 meters) above the Earth, we can substitute this value into the equation as the velocity "v". The mass of the Earth (mE) is not necessary for this calculation.
r = GmE / (200,000)^2
Since the problem does not provide the constant "G", it is assumed that you have the value to substitute in.
Once you have the appropriate value for "G", you can calculate the radius of the satellite's orbit (r) by substituting the known values into the equation and performing the calculations.
To find the radius of the satellite's orbit, we need to subtract the height of the satellite from the radius of the Earth.
Given:
- Height of the satellite above Earth's surface = 200 kilometers = 200,000 meters
- Radius of Earth = 6,380,000 meters
We can calculate the radius of the satellite's orbit as follows:
Radius of satellite's orbit = Radius of Earth + Height of the satellite
= 6,380,000 meters + 200,000 meters
So, the radius of the satellite's orbit is 6,580,000 meters.
From your given information,
r = 6,380,000 + 200,000 = 6,580,000 m or
....6,380 + 200 = 6,580 km.
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, ~1.40766x10^16 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 = 1.40766x10^16/[(3963+250)x5280] = 1.40766x10^16/22,244,640 = 25,155 fps. (17,147 mph.) Since velocity is inversely proportional to r, the higher you go, the smaller the required orbital velocity.
Insert your data to derive your orbital velocity.