Calculate the first space velocity at the launch from the surface of Jupiter, if it is known that one of its satellites rotates in a nearly circular orbit of radius $r = 10^{6} km$ with period $T = 7.1 days$. The radius of Jupiter is $R = 7 \cdot 10^{4} km$.
Decision:
A satellite orbiting an orbit of radius $r$ with period $T$ moves with centripetal acceleration
$a_{centripetal}=\frac{v^{2}}{r}= \frac{4 \pi^{2}r}{T^{2}}$. (1)
This acceleration is produced by the gravitational force from the planet. It follows from Newton's 2nd law of motion that $a_{centripetal}$ that
$ a_{centripetal} = \frac{GM}{r^{2}}$, (2)
where $M$ is the mass of Jupiter. Equating the right-hand sides of (1) and (2), we find:
$GM= \frac{4 \pi^{2}r^{3}}{T^{2}}$. (3)
The first space velocity $v_{1}$ is the velocity of the satellite in an orbit near a planet of radius $R$, i.e., at $r = R$. It must satisfy the condition
$\frac{v^{2}_{1}}{R} = a_{centripetal}= \frac{GM}{R^{2}}$. (4)
Replacing in (4) the product of $GM$ by the right-hand side of (3) and then solving the resulting equation with respect to $v_{1}$, we find:
$v_{1}=\frac{2 \pi r}{T} \sqrt{\frac{r}{R}} \approx 3.32 \cdot 10^{2} \frac{km}{day}=1.38 \cdot \frac{km}{h}=38.3 \frac{km}{s}$
Decision:
A satellite orbiting an orbit of radius $r$ with period $T$ moves with centripetal acceleration
$a_{centripetal}=\frac{v^{2}}{r}= \frac{4 \pi^{2}r}{T^{2}}$. (1)
This acceleration is produced by the gravitational force from the planet. It follows from Newton's 2nd law of motion that $a_{centripetal}$ that
$ a_{centripetal} = \frac{GM}{r^{2}}$, (2)
where $M$ is the mass of Jupiter. Equating the right-hand sides of (1) and (2), we find:
$GM= \frac{4 \pi^{2}r^{3}}{T^{2}}$. (3)
The first space velocity $v_{1}$ is the velocity of the satellite in an orbit near a planet of radius $R$, i.e., at $r = R$. It must satisfy the condition
$\frac{v^{2}_{1}}{R} = a_{centripetal}= \frac{GM}{R^{2}}$. (4)
Replacing in (4) the product of $GM$ by the right-hand side of (3) and then solving the resulting equation with respect to $v_{1}$, we find:
$v_{1}=\frac{2 \pi r}{T} \sqrt{\frac{r}{R}} \approx 3.32 \cdot 10^{2} \frac{km}{day}=1.38 \cdot \frac{km}{h}=38.3 \frac{km}{s}$