Calculating the velocity of a circular 200km orbit around Kerbin

[AlexGodbehere]

I'm studying a space systems module at university, and thought I'd try and apply some of the things I've learnt to spacecraft in KSP.

I'm using the formula:

V = sqrt(MG/r) to work out the velocity of the satellite when at a 200km orbit around Kerbin. The values I'm using are:

MG = 3.5316000×10^12 m3/s2 (KSP Wiki)

r = 800 (600km radius + 200km orbit)

I'm obtaining a value of 66441. When I launch a satellite, the surface speed is 1.868km/s.

Which is way way out. Its not even out because of units, its just plain wrong. Has anyone got any ideas?

Edited December 24, 2013 by AlexGodbehere

[Stealth2668]

Clearly you broke physics! >:{

No idea though

[Brotoro]

R = 800,000 meters

[ttb]

For a start, use 800000m instead of 800km for kerbin radius + altitude.

The correct formula is v = R sqrt(g/(R+h)) so we can plug in some known values

v = 600000 sqrt(9.807/800000) = 2425 m/s

Hope thus helps

ttb

[sanoj688]

Why would you look at the surface speed anyway?

[AlexGodbehere]

Why would you look at the surface speed anyway?

What speed would I look at using MechJeb?

[ttb]

Orbital speed I think

[Brotoro]

What speed would I look at using MechJeb?

Just the speed on the an ball (orbit). Your ground speed will read too low because the surface of Kerbin is moving toward the east as the planet rotates.

[Supernovy]

2425 m/s is much too high, LKO is 2300ish and everything above that will be slower. Let's try going to first principles:

The force of gravity must be the same as the centripetal force, GMm/r^2 = mv^2/r , then cancelling: GM/r = v^2. Taking the square root gives the equation in the OP, sqrt(GM/r) = v.

Taking the gravitational parameter from the wiki (again, same as in OP) and 800000 metres radius, sqrt(3.532x10^12 / 8x10^5) = 2.101x10^3 m/s.

2101 metres per second. I think at 240ish kilometres it hits 2 km/s exactly, so this seems like a sensible result.

[AlexGodbehere]

2425 m/s is much too high, LKO is 2300ish and everything above that will be slower. Let's try going to first principles:

The force of gravity must be the same as the centripetal force, GMm/r^2 = mv^2/r , then cancelling: GM/r = v^2. Taking the square root gives the equation in the OP, sqrt(GM/r) = v.

Taking the gravitational parameter from the wiki (again, same as in OP) and 800000 metres radius, sqrt(3.532x10^12 / 8x10^5) = 2.101x10^3 m/s.

2101 metres per second. I think at 240ish kilometres it hits 2 km/s exactly, so this seems like a sensible result.

Thanks. Yeah, using that equation with 800000 instead of 800 works. Teaches me for not paying attention to my units. Cheers everyone.

[Luckless]

Thanks. Yeah, using that equation with 800000 instead of 800 works. Teaches me for not paying attention to my units. Cheers everyone.

Well, now you can feel like real life rocket scientists and engineers! They've ruined more than one real world multimillion dollar space craft and had near misses thanks to unit mix-ups.

[ttb]

2425 m/s is much too high, LKO is 2300ish and everything above that will be slower. Let's try going to first principles:

The force of gravity must be the same as the centripetal force, GMm/r^2 = mv^2/r , then cancelling: GM/r = v^2. Taking the square root gives the equation in the OP, sqrt(GM/r) = v.

Taking the gravitational parameter from the wiki (again, same as in OP) and 800000 metres radius, sqrt(3.532x10^12 / 8x10^5) = 2.101x10^3 m/s.

2101 metres per second. I think at 240ish kilometres it hits 2 km/s exactly, so this seems like a sensible result.

Lol oops, back to the drawing board for me thanks for the heads up

ttb