Question

A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth’s mass is 6 × 1024 kg and its radius is 6.4 × 106 m.

The distance of the satellite from Earth, written in standard notation, is _____?_____m.

Answer 1

When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.

Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration

$\frac{ M_{E}*G }{ ( r_{E} )^{2} } =$ω^2r

where

$M_{E}$ = mass of earth
G = gravitational constant = 6.6742 x 10-11 m3 s-2 kg-1
$r_{E}$ = radius of earth
ω = angular velocity
r = radius of orbit

To convert to angular velocity:

Tangential velocity = rω
ω = 5000/r

Then,

$\frac{ (6 \ x \ 10^{24}) *(6.6742 \ x \ 10^{-11} ) }{ ( 6.4 \ x \ 10^{6})^{2} }= ( \frac{5000}{r} )^{2} r$

r = 2557110.465 m

Therefore, the distance of the centers of the earth and the satellite is 2.6 x 10^6 m.

Answer 2

Answer:

The answer is 9,600,000 on edgen