Question

a synchronous satellite circles the earth eastward above equator once every 24h and stays over the same spot on the earth because the earth is rotating at the same rate. what is the velocity of the synchronous satellite?

9800 m/s

4300 m/s

3100 m/s

2400 m/s

Answer 1

Answer on Question 56222, Physics, Mechanics, Relativity

Question:

A synchronous satellite circles the Earth eastward above equator once every $24h$ and stays over the same spot on the Earth because the Earth is rotating at the same rate. What is the velocity of the synchronous satellite?

a) $9800 \, m/s$

b) $4300 \, m/s$

c) $3100 \, m/s$

d) $2400 \, m/s$

Solution:

Let's first find the orbital radius of the synchronous satellite. When the satellite orbits the Earth the centripetal force acts on it:

$$F_c = \frac{m_{sat} v^2}{R_{sat}},$$

where, $m_{sat}$ is the mass of the satellite, $v$ is the orbital speed of the satellite and $R_{sat}$ is the orbital radius of the satellite.

From the other hand, the gravitational force attracts the satellite towards the Earth, and we can write:

$$F_{grav} = G \frac{m_{sat} M_E}{R_{sat}^2},$$

where, $G = 6.67 \cdot 10^{-11} \frac{Nm^2}{kg^2}$ is the gravitational constant, $M_E = 5.98 \cdot 10^{24} kg$ is the mass of the Earth.

Since, $F_c = F_{grav}$, we obtain:

$$\frac{v^2}{R_{sat}} = G \frac{M_E}{R_{sat}^2}.$$

Because the satellite travels around the entire circumference of the circle which is $2\pi R_{sat}$ in the period $T$, this means that the orbital speed must be $v = \frac{2\pi R_{sat}}{T}$. Substituting the expression for the orbital speed into the last equation we get:

$$\frac {\left(\frac {2 \pi R _ {s a t}}{T}\right) ^ {2}}{R _ {s a t}} = G \frac {M _ {E}}{R _ {s a t} ^ {2}}.$$

Finally, after simplification we get the formula for the orbital radius of the synchronous satellite:

$$\begin{array}{l} R _ {s a t} = \sqrt [ 3 ]{\frac {G M _ {E} T ^ {2}}{4 \pi^ {2}}} = \sqrt [ 3 ]{\frac {6 . 6 7 \cdot 1 0 ^ {- 1 1} \frac {N m ^ {2}}{k g ^ {2}} \cdot 5 . 9 8 \cdot 1 0 ^ {2 4} k g \cdot (2 4 \cdot 3 6 0 0 s) ^ {2}}{4 \cdot (3 . 1 4) ^ {2}}} \\ = 4. 2 2 6 \cdot 1 0 ^ {7} m. \\ \end{array}$$

Then, we can find the orbital speed of the synchronous satellite:

$$v = \frac {2 \pi R _ {s a t}}{T} = \frac {2 \cdot 3 . 1 4 \cdot 4 . 2 2 6 \cdot 1 0 ^ {7} m}{2 4 \cdot 3 6 0 0 s} = 3 0 7 2 \frac {m}{s} \sim 3 1 0 0 \frac {m}{s}.$$

Answer:

c) $3100 \, \text{m/s}$

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