To understand how the velocity of a satellite changes when it moves from an orbital radius \( r \) to a new orbital radius of \( 5r \), we can use the principles of orbital mechanics.
The gravitational force acting on the satellite provides the necessary centripetal force for it to maintain its circular orbit. The gravitational force \( F \) acting on the satellite is given by:
\[ F = \frac{G M m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.
The centripetal force required to keep the satellite in orbit is given by:
\[ F = \frac{m v^2}{r} \]
where \( v \) is the orbital velocity of the satellite at radius \( r \).
Equating the two expressions for force gives:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]
We can simplify this by canceling \( m \) and rearranging the equation:
\[ v^2 = \frac{G M}{r} \]
Taking the square root to find the velocity:
\[ v = \sqrt{\frac{G M}{r}} \]
Now, when the satellite is moved to an orbital radius of \( 5r \), we can use the same formula to find the new velocity \( v' \):
\[ v' = \sqrt{\frac{G M}{5r}} \]
We can relate the new velocity \( v' \) to the original velocity \( v \):
\[ v' = \sqrt{\frac{G M}{5r}} = \sqrt{\frac{1}{5}} \cdot \sqrt{\frac{G M}{r}} = \sqrt{\frac{1}{5}} v \]
Thus, the velocity \( v' \) at the new orbital radius \( 5r \) is:
\[ v' = \frac{v}{\sqrt{5}} \]
In summary, when a satellite is moved from an orbital radius \( r \) to an orbital radius \( 5r \), its new orbital velocity is \( \frac{v}{\sqrt{5}} \). This means that the new velocity is approximately 0.447 times the original velocity.