Answer:
2.83
Explanation:
Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:
(1)
with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):
(2)
Because in the right side of the equation (2) we have only constant quantities, that implies the ratio 
is constant for all the planets orbiting the same sun, so we can said that:
<span>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</span>
Answer:
gkcfgggxg hhfdc ghgffgccvvc
bjvhgf
Explanation:
gjhghujj bhhgghj
