72.7k views
4 votes
Calculate the speed of a satellite moving in a stable circular orbit about the earth at a height of 2600 km from the surface of the earth (the mass of the earth is 5.97Ã1024 kg and the radius of the earth is 6.38Ã106 m).

User JIV
by
8.4k points

1 Answer

1 vote
6.66 km/s The velocity of a body in any given point in an orbit is v = sqrt(u(2/r - 1/a)) where u = Standard gravitational parameter r = radius at which speed is to be calculated a = length of semi-major axis Since we're using a circular orbit, the equation can be simplified to v = sqrt(u(2/r - 1/r)) = sqrt(u(1/r)) = sqrt(u/r) u is the product of the body's mass and the gravitational constant. So u = 6.67408 x 10^-11 m^3/(kg s^2) * 5.97 x 10^24 kg = 3.9844 x 10^14 m^3/s^2 The radius of the orbit will be the sum of earth's radius and the satellite's altitude. So r = 6.38x10^6 m + 2.60 x 10^6 m = 8.98 x 10^6 m Now plugging in the calculated values of u and r into the equation above gives v = sqrt(u/r) = sqrt(3.9844 x 10^14 m^3/s^2 / 8.98 x 10^6 m) = sqrt(4.43697x10^7 m^2/s^2) = 6661 m/s = 6.661 km/s Since we only have 3 significant figures in our data, round the result to 3 figures, giving 6.66 km/s For orbital calculations, it's far better to be provided GM instead of M for the body being orbited. Reason is that GM is known to more accuracy than either G or M. Taking the earth for example, the value GM is known to 10 significant digits, whereas G is only known to 6 significant digits and M is known to only 5 significant digits. The reason the accuracy for GM is known so much more precisely is because of extended observations of satellites in orbit.
User Lazar Vuckovic
by
8.1k points