The hubble space telescope orbits 569 km above earths surface. given that earth's mass is 5.97 x 10 to the 24th kg and its radius is 6.38 x 10 to the 6th m whats is HST's tangential speed

We can use an equation to find the gravitational force exerted on the HST.

F = GMm / r² G is the gravitational constant, M is the mass of the Earth, m is the mass of the HST, r is the distance to the center of the Earth. This force, F, provides the centripetal force for the HST to move in a circle.

The equation we use for circular motion is: F = mv^2 / r, m is the mass of the HST, v is the tangential speed, r is the distance to the center of the Earth. Now we can equate these two equations to find v.

mv² / r = GMm / r² v² = GM / r v = √{GM / r } v = √{ (6.67 x 10^{-11}) (5.97 x 10^{24}) / 6,949,000 m} v = 7570 m/s which is equal to 7.570 km/s

HST's tangential speed is 7570 m/s or 7.570 km/s