Calculating Orbital Velocity for a Satellite Orbiting Earth

[.:Endeavour:.]

Can you find the orbital velocity of a planet which a satellite has to obtain in order to revolve around the planet with this formula:

v = [tex]\sqrt{\frac{Gm_E}{r_E}}[/tex]

G = gravitational constant 6.67*10^-11
m_E = Mass of a planet
r_E = the radius of the orbit

r_E = h + r
h = height of the orbit
r = the radius of a plant from the center of the planet to its surface

Lets see, if you want to find the orbital velocity that a satellite has to get in order to orbit the Earth in Low Earth Orbit (346.9 km).

m_E = 5.98*10²⁴ kg
r_E = h + r
h = 346,900 meters (346.9 km)
r = 6.38*10⁶ meters

v = [tex]\sqrt{\frac{(6.67*10^-^1^1)(5.98*10^2^4 kg))}{6,726,900}}[/tex]
v = 7,700.27 m/s

In order to obtain an orbit at an altitude of 346.9 km you will need to get a velocity of 7,700.27 m/s. Is this right, if not please correct me. Thank you for your time.

 

[D H]

In order to obtain an orbit at an altitude of 346.9 km you will need to get a velocity of 7,700.27 m/s. Is this right, if not please correct me. Thank you for your time.

That is basically correct. To pick a nit: This result, 7,700.27 m/s, has too much precision. You used a three-figure value for both G and for the radius of the Earth. Your result has thusly at most three significant figures, so saying something down to centimeter per seconds is incorrect. This result also ignores that orbits do not have to be circular, or that there is no such thing as a perfectly circular orbit, particularly in low-Earth orbit.

If you had said 7.7 kilometers per second instead of 7,700.27 m/s I wouldn't have picked so many nits.

 

[.:Endeavour:.]

That is basically correct. To pick a nit: This result, 7,700.27 m/s, has too much precision. You used a three-figure value for both G and for the radius of the Earth. Your result has thusly at most three significant figures, so saying something down to centimeter per seconds is incorrect. This result also ignores that orbits do not have to be circular, or that there is no such thing as a perfectly circular orbit, particularly in low-Earth orbit.

If you had said 7.7 kilometers per second instead of 7,700.27 m/s I wouldn't have picked so many nits.

The velocity is an approximate, but like you said it isn't exact. I agree that the 7,700.27 m/s does look too precise. I was using my book to refer on what the units to use for the velocity
which gave m/s.

For a rocket to reach 7.7 km does it have to reach that velocity in a period of time or not? Because most rockets like the Space Shuttle reach LEO orbit around 8 to 10 minutes. Just kind of curious there.