mass * centripetal acceleration = mass * gravity at that height
You might as well assume g = 9.8 I suppose because 350 km is pretty small compared to earth radius.
In any case you see the mass does not matter. A feather would stay in orbit beside you.
m v^2/r = m g
v^2 = gr
r = 350,000 + radius of earth
where earth radius = 6,380,000 meters
to be more accurate use
g at height = G Me/r^2 = 6.67*10^-11 * 5.98*10^24/r^2
Calculate the speed needed by a spacecraft to maintain a circular orbit at about 350 km above Earth’s surface.
(it's a project where we create our own spaceship. let's say the craft is 500kg)
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Not sure about the formula to find this. All I know is that it may have something to do with the escape velocity of earth, which is about 1.1x10^4m/s.
2 answers
The orbital speed for low Earth orbit is 71% of the escape velocity. You are right; there is a connection.
You can derive it using the formula:
V^2/R = GM/R^2
V = sqrt(GM/R)
Where R is the distance from the center of the Earth, which in this case is 6371 + 350 = 6721 km. G is the universal constant of gravity,
6.673*10−11 N m^2 kg^−2, and M is the mass of the Earth, 5.98 × 10^24 kg.
V = 7.70*10^3 m/s
You can derive it using the formula:
V^2/R = GM/R^2
V = sqrt(GM/R)
Where R is the distance from the center of the Earth, which in this case is 6371 + 350 = 6721 km. G is the universal constant of gravity,
6.673*10−11 N m^2 kg^−2, and M is the mass of the Earth, 5.98 × 10^24 kg.
V = 7.70*10^3 m/s
1 answer