Question

the period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 4R is?

Answer 1

Answer on Question #42595, Physics, Mechanics | Kinematics | Dynamics

Question.

The period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $4R$ is?

Solution.

Newton's law of universal gravitation:

$$F = G \frac{Mm}{R^2}$$

$F$ is the force between bodies M and m, the force with which body $M$ acts on body $m$;

$M$ is a mass of planet;

$m$ is a mass of satellite;

$R$ is a distance between $M$ and $m$;

$G = 6.67 * 10^{-11} \frac{N \cdot m^2}{kg^2}$ is a gravitational constant.

Newton's second law for the satellite:

$$F = m a$$

$a$ is the centripetal acceleration.

$$a = \frac{v^2}{R} = \omega^2 R$$ $$v = \omega R$$

$v$ is linear velocity;

$\omega$ is the angular velocity.

$$\omega = \frac{2\pi}{T}$$

$T$ is a period of motion.

So,

$$a = \frac{v^2}{R} = \omega^2 R \rightarrow a = \frac{4\pi^2}{T^2} R$$ $$F = m a \rightarrow G \frac{Mm}{R^2} = m \frac{4\pi^2}{T^2} R$$

Therefore,

$$T ^ {2} = \frac {4 \pi^ {2}}{G M} R ^ {3}$$ $$T = \sqrt {\frac {4 \pi^ {2}}{G M}} R ^ {3 / 2}$$

Thus, if $R_0 \to 4R_0$ :

$$T _ {0} = \sqrt {\frac {4 \pi^ {2}}{G M}} R _ {0} ^ {3 / 2}$$ $$T = \sqrt {\frac {4 \pi^ {2}}{G M}} R ^ {3 / 2} = \sqrt {\frac {4 \pi^ {2}}{G M}} (4 R _ {0}) ^ {3 / 2} = \sqrt {\frac {4 \pi^ {2}}{G M}} R _ {0} ^ {3 / 2} \cdot 4 ^ {3 / 2} = \sqrt {\frac {4 \pi^ {2}}{G M}} R _ {0} ^ {3 / 2} \cdot 8 = 8 T _ {0}$$

So, if $R_0 \to 4R_0$ , then $T \to 8T_0$

Answer.

8T

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