Question

What is the orbital radius and speed of a synchronous satellite which orbits the earth once every 24 hours. Take G = 6.67 x 10 raise to power -11 Nm/Kg2, mass of the earth is 5.98 x 10 raise to power 24kg.

Answer 1

Answer on Question #55220, Physics Mechanics Kinematics Dynamics

What is the orbital radius and speed of a synchronous satellite which orbits the earth once every 24 hours. Take $\mathrm{G} = 6.67 \times 10$ raise to power -11 Nm/Kg2, mass of the earth is 5.98 x 10 raise to power 24kg.

Solution

Fig.1

Consider a satellite with mass $M_{s}$ orbiting a central body with a mass of mass $M_{E}$ . The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object (in our example body is Earth). If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship

$$F _ {n e t} = M _ {S} \cdot \frac {v ^ {2}}{R} = M _ {S} \cdot \frac {\left(2 \pi R / T\right) ^ {2}}{R} = \frac {4 \pi^ {2} R M _ {S}}{T ^ {2}}$$

where $R$ is the orbital radius; $\nu$ is the speed of satellite

This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented as

$$F _ {\text {g r a v}} = \frac {G \cdot M _ {E} \cdot M _ {S}}{R ^ {2}}$$

Since $F_{\text{grav}} = F_{\text{net}}$ , the above expressions for centripetal force and gravitational force can be set equal to each other. Thus,

$$\frac {4 \pi^ {2} R M _ {S}}{T ^ {2}} = \frac {G \cdot M _ {E} \cdot M _ {S}}{R ^ {2}} \Rightarrow R = \sqrt [ 3 ]{\frac {G \cdot M _ {E} \cdot T ^ {2}}{4 \pi^ {2}}}$$

Then

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

The speed is given by Eq.(4)

$$v = 2 \pi R / T = 2 \cdot 3. 1 4 \cdot 4 2 2 0 0 \mathrm {k m} / (2 4 \cdot 3 6 0 0 s) = 3. 0 7 \mathrm {k m} / s$$

$2\pi R$ is the circumference of the orbit.

Answer: $R = \sqrt[3]{\frac{G \cdot M_E \cdot T^2}{4\pi^2}} = 42200km$ ; $v = 2\pi R / T = 3.07km / s$

http://www.AssignmentExpert.com/