# Orbital Velocity Formula with solved numerical problem

Last updated on February 12th, 2022 at 03:08 pm

Here in this post, we will list down different forms of the **Orbital velocity formula** or equation. We will present here 3 sets of formulas including the near-orbit one. After this, we will also see how to use these formulas to solve numerical problems.

**Orbital Velocity Formula – 3 sets of formulas are listed below:**

**Orbital Velocity Formula Set 1:**

**V = [(GM)/r] ^{1/2}** ……………. (1)

( here G = gravitational Constant,

M is the mass of the central body i.e. for the formula of the orbital velocity of the earth’s satellite, M denotes the mass of the earth,

Here R = radius of the earth, h = height above the earth’s surface, And

**r = R +h**)

**Orbital Velocity Formula Set 2:**

**V = R . (g/r) ^{1/2} ……. (2)**

(Here R = radius of the earth, h = height above the earth’s surface, And

**r = R +h**)

**Orbital Velocity Near Orbit equation Set 3:** [ for near-orbit satellite]

**(2 expressions for Near Orbit)****V = [(GM)/R] ^{1/2}** ———–(3)

**V**=

_{orbital}**(gR)**here, R =radius of the earth

^{1/2}……….(4)**Formula sets (Graphical Representation)**

**Derivation of the above-listed formula**

**Sample Numerical problem** based on **Orbital velocity equation **– with solution

Q) **Assume that a satellite orbits Earth 225 km above its surface. Given that the mass of Earth is 5.97 x 10**^{24}** kg and the radius of Earth is 6.38×10**^{6}** m, what is the satellite’s orbital speed?**

** Solution:**** **h =2.25×10^{5} m (height of the satellite’s orbit)

r_{E} =6.38×10^{6} m (Earth’s radius)

m_{E} =5.97×10^{24} kg (Earth’s mass)

G = 6.67×10^{-11} Nm^2/kg^2

Orbital velocity = V =?

We get the orbital radius r by adding the height of the satellite’s orbit to Earth’s radius.

**r** =h +r_{E} =2.25×10^{5} m + 6.38×10^{6} m = 6.61×10^{6} m

So, Orbital speed.

V = √[(G m_{E})/r] = [G m_{E})/r ]^{1/2 } = 7.76 x 10^{3} m/s

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**Derivation of the above-listed formula**

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