# What is the orbital velocity at a height of 300 km above the surface of the earth? (The mass of...

## Question:

What is the orbital velocity at a height of 300 km above the surface of the earth? (The mass of Earth is {eq}6 \times 10^{24} {/eq} kg and its radius is {eq}6.4 \times 10^6 {/eq} m.)

## Orbital Motion and Satellite:

Any object at the orbit around any planet must maintain a certain velocity to keep itself rotating around the planet on a stable orbit. This velocity is called an orbital velocity and depends on the mass of the planet and the distance from the object to the planet.

Any satellite can only remain in any orbit when centripetal force acting on satellite is balanced by gravitational force Earth and satellite.

The motion of a satellite can be described as follows:

{eq}\dfrac {mv^2}{R_E + h} = \dfrac {GM_Em}{(R_e + h)^2} {/eq}

Here

• {eq}h = 300 \ km {/eq} is the height of the satellite;
• {eq}G = 6.67\times 10^{-11} \ N\cdot m^2/kg^2 {/eq} is the universal gravitational constant;

Solving for the velocity, we obtain:

{eq}v = \sqrt{ \dfrac{GM_E}{R_E + h}} {/eq}

Calculating, we get:

{eq}v = \sqrt{ \dfrac{6.67\times 10^{-11} \ N\cdot m^2/kg^2 \times 6\times 10^{24} \ kg}{6.4\times 10^6 \ m + 3\times 10^5 \ m}} \approx 7729 \ m/s {/eq} 