Earth has a mass of 5.97 x 1024kg, and a mean radius of 6.38 x 10m. what would be the orbita
satellite in orbit 1.44 x 108 m above earth?
a. v, = 630m/s,
c. v, - 1630m/s,
t = 2.78 x 10 s
t=5.78 x 10s
b. v = 1820m/s,
d. v, - 1260m/s,
t = 6.78 x 106 s
t=5.78 x 10's
select the best answer from the choices provid

v = 1630 m/s

T = 5.78 x 10^5 s

Explanation:

The tangential speed of the satellite can be found by requiring that the gravitational force on the satellite is equal to the centripetal force:

where

G is the gravitational constant

M=5.97 x 1024kg is the Earth's mass

m is the satellite's mass

is the Earth's radius

is the altitude of the satellite

v is the speed of the satellite

Solving for v,

And the period of the orbit is equal to the ratio between the distance covered during one revolution (the circumference of the orbit) and the speed:

So the correct answer is

v = 1630 m/s

T = 5.78 x 10^5 s