r=6.4E6+.6E6 m
Does that do it for you?
The Hubble Space Telescope orbits at an altitude of 600 km above the Earth's surface. Assuming it is in a stable circular orbit, what is its velocity? For reference, the Earth's radius is 6400 km and Earth's mass is 6 x 10^24 kg.
The formula to solve for circular velocity is v(circ) = the square root of (GM/r)
The answer is 7500 m/s but I'm not sure how to get to this answer from this equation
2 answers
What you want to do is add the altitude to the earth's radius to get the orbital radius. In your case, you'd add 600 km to 6400 km to get 7000 km. Use 7000 km as your r value and plug it into the formula. When you're trying to find r, measure the distance from the center of the earth (or planet I guess) to the object.
Vcirc= square root of (GM/r)
Here's a similar example:
G = 6.67 × 10^−11 m3 / (kg s2)
MEarth = 5.98 × 10^24 kg
rEarth = 6.4 × 10^6 m
altitude of roughly 6 × 10^5 meters
Add altitude of object and radius of earth to get orbital radius:
r= 7 x 10^6
Use this r value in the formula: Vcirc= square root of (GM/r)
In my example, you'll get ~7579 m/s
Vcirc= square root of (GM/r)
Here's a similar example:
G = 6.67 × 10^−11 m3 / (kg s2)
MEarth = 5.98 × 10^24 kg
rEarth = 6.4 × 10^6 m
altitude of roughly 6 × 10^5 meters
Add altitude of object and radius of earth to get orbital radius:
r= 7 x 10^6
Use this r value in the formula: Vcirc= square root of (GM/r)
In my example, you'll get ~7579 m/s
1 answer