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).

Respuesta :

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.
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