A satellite flies around the earth in a circular orbit with radius $R = 25600 km$. How many times is the satellite's speed different from the first space velocity?
Decision:
The centripetal force for a satellite flying in a circular orbit is the gravitational force $F = \frac{GmM}{R^{2}}$, where $G$ is the gravitational constant, $m$ and $M$ are the masses of the satellite and the Earth, respectively. This force creates a centripetal acceleration $\frac{v^{2}}{R}$, where $v$ is the linear velocity of the satellite. According to Newton's II law $(F=ma)$
$\frac{GmM}{R^{2}} = \frac{mv^{2}}{R}$. (1)
The first space velocity $v_{1}$ is the velocity of a satellite traveling in a circular orbit whose radius is the same as the radius of the of the Earth. Thus, the equality
$G \frac{mM}{R^{2}_{e}} =m \frac{v^{2}_{1}}{R_{e}}$, (2)
Dividing the equality (1) by the equality (2) and extracting the square root, we get:
$\frac{v}{v_{1}} = \sqrt{ \frac{R_{e}}{R}} = \frac{1}{2}$. (3)
Decision:
The centripetal force for a satellite flying in a circular orbit is the gravitational force $F = \frac{GmM}{R^{2}}$, where $G$ is the gravitational constant, $m$ and $M$ are the masses of the satellite and the Earth, respectively. This force creates a centripetal acceleration $\frac{v^{2}}{R}$, where $v$ is the linear velocity of the satellite. According to Newton's II law $(F=ma)$
$\frac{GmM}{R^{2}} = \frac{mv^{2}}{R}$. (1)
The first space velocity $v_{1}$ is the velocity of a satellite traveling in a circular orbit whose radius is the same as the radius of the of the Earth. Thus, the equality
$G \frac{mM}{R^{2}_{e}} =m \frac{v^{2}_{1}}{R_{e}}$, (2)
Dividing the equality (1) by the equality (2) and extracting the square root, we get:
$\frac{v}{v_{1}} = \sqrt{ \frac{R_{e}}{R}} = \frac{1}{2}$. (3)