Question

An astronaut orbiting the earth is preparing to dock with a westar vi satellite. The satellite is in a circular orbit 500 km above the earth's surface, where the free-fall acceleration is 8.25 m/s2. Take the radius of the earth as 6400 km. Determine the speed of the satellite.

Answer 1

As the satellite is in free fall, therefore we use the formula

$v = \sqrt{r g}$

Here, v is the orbital speed, r is radius of Earth plus satellite is in a circular orbit 500 km above the earth's surface and g is the satellite free-fall acceleration    .

Given   $r = 6400 \ km + 500 \ km = 6900 \ km$ and $g = 8.25 \ m/s^2$

Substituting these values in above equation we get,

$v = \sqrt{(6900 \times 10 ^3 \ m ) (8.25 \ m/s^2) } = 7.54 \times 10 ^3 \ m/s$

Thus, the speed of satellite in  order to maintain the  orbit is $7.54 \times 10 ^3 \ m/s$

Answer 2

Answer:

Speed of satellite =7.59×10^3m/s

Explanation:

Formular for speed of satellite is given by:

V= Sqrt(Gm)/(RE + r))

Where G= gravitational constant =6.67×10^-11m^3kgs

m is mass of the earth =5.97×10^24kg

V= sqrt[(6.67×10^-11)(5.97×10^24)/(6900×10^3)]

V=Sqrt(3.98×10^14)/(6.9×10^6)

V= sqrt(57681159.42)

V= 7594.81m/s

V= 7.59×10^3m/s