Question

4. Find the orbital speed of a satellite in a circular orbit 3×10^7 m above the surface of the earth. Me= 5.97 x10^24 kg, Re =6.37 x 10^6 m)

Answer 1

v = 3308.86 m/s

Step-by-step explanation

The centripetal force acting on a satellite moving at an orbital speed v, with radius R, and mass m is,

`F_c=m\cdot(v^2)/(R)`

Also, the force of attraction between two bodies of masses M and m, separated by a distance R is,

`F_g=G\cdot(M\cdot m)/(R^2)`

These two forces are the same since no other force is acting on the satellite,

`m\cdot(v^2)/(R)=G\cdot(M\cdot m)/(R^2)`

Solving for v we have,

`v=\sqrt[]{(G\cdot M)/(R)}`

Where:

• G is the gravitational constant, 6.67x10⁻¹¹ Nm²/kg²

,

• M is the mass of the larger body, in this case, the Earth, 5.97x10²⁴ kg

,

• R is the radius of the orbit

In this case, we know that the satellite's orbit is 3x10⁷m above the surface of the Earth, so the orbit's radius is the sum of this distance and the radius of the Earth,

So the orbit's radius is,

`R=6.37*10^6m+3*10^7m=3.637*10^7m`

Now we can input these values into the equation above to find the orbital speed,

`v=\sqrt[]{(6.67*10^(-11)Nm^2/kg^2\cdot5.97*10^(24)kg)/(3.637*10^7m)}\approx3308.86m/s`

Hence, the orbital speed of the satellite is 3308.86 m/s.