The hubble space telescope orbits Earth 612 km above Earth's surface, what is the period of the tele's orbit?

Answer 1

You need a centrifugal force.
F=mr*(omega)^2
The question is not completed.
F=mg, you need the g at 612 km above the Earth's surface.

Answer 2

Any object in low earth orbit will have an orbital period of about 90 minutes. For a more exact value, you will need to use the universal gravitational constant, the mass of the earth, and the radius of the orbit as taken from the earth's center. Alternatively, use the gravitational acceleration at the earth's surface, calculate the reduction in that due to the orbit's distance above the surface, and use that force to balance the centrifugal force arising from the satellite's inertia. The latter computation will be simpler because it avoids dealing with huge numbers.

Answer 3

Any merchandise in low earth orbit ought to have an orbital era of roughly ninety minutes. For a greater ideal fee, you will could desire to persist with the trouble-free gravitational consistent, the mass of the earth, and the radius of the orbit as taken from the earth's center. having reported that, use the gravitational acceleration in the international's floor, calculate the relief in that because of utilising orbit's distance above the outdoors, and use that rigidity to stability the centrifugal rigidity arising from the satellite tv for pc's inertia. The latter computation could be much less complicated because of utilising certainty it avoids coping with extensive numbers.

Answer 4

m*r*omega^2 = m*g
assuming gravity remains constant at at distance.
r = radius of earth+612 km
solve for omega.