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

ANSWER

v = 3308.86 m/s

EXPLANATION

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

$F_c=m\cdot\frac{v^2}{R}$

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

$F_g=G\cdot\frac{M\cdot m}{R^2}$

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

$m\cdot\frac{v^2}{R}=G\cdot\frac{M\cdot m}{R^2}$

Solving for v we have,

$v=\sqrt[]{\frac{G\cdot M}{R}}$

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,

$R=6.37\times10^6m+3\times10^7m=3.637\times10^7m$

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

$v=\sqrt[]{\frac{6.67\times10^{-11}Nm^2/kg^2\cdot5.97\times10^{24}kg}{3.637\times10^7m}}\approx3308.86m/s$

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