Question

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius is 6.38 × 106 m, what is HST’s tangential speed?

Answer 1

We can use an equation to find the gravitational force exerted on the HST. F = GMm / r^2 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. mv^2 / r = GMm / r^2 v^2 = GM / r v = sqrt{GM / r } v = sqrt{(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

Answer 2

Answer:

Just type 7,570

Explanation: