# 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.

## Answer and Explanation:

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}

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from Physics: High School

Chapter 8 / Lesson 16