Calculations of orbital velocity — lesson. Science State Board, Class 9.

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Formula of orbital velocity:

The orbital velocity is calculated using the formula,

$v = \sqrt{\frac{GM}{(R + h)}}$

Where,

$G$ is the Gravitational constant (

$6.673 \times 10^{–11}$

$Nm^2kg^{-2}$ ),

$M$ is the mass of the Earth (

$5.972 \times 10^{24}$

$kg$ ),

$R$ is the radius of the Earth (

$6371\ km$ ),

$h$ is the height of the satellite from the Earth's surface

Example:

Calculate the orbital velocity of a satellite orbiting at an altitude of

$500\ km$ above Earth?

$G$

$=$

$6.673 \times 10^{–11}$

$Nm^2kg^{-2}$

$M$

$=$

$5.972 \times 10^{24}$

$kg$

$R$

$=$

$6731000\ m$

$h$

$=$

$500000\ m$

The formula is given as

$v = \sqrt{\frac{GM}{(R + h)}}$

On substituting the given values, we get

$\begin{array}{l} v = \sqrt{\frac{6.673 \times 10^{- 11} \times 5.972 \times 10^{24}}{(6731000 + 500000)}} \\ = \sqrt{\frac{39.851156 \times 10^{13}}{(7231000)}} \\ = 7613 m s^{- 1} \\ = 7. 613 km s^{- 1} \end{array}$

Thus, the orbital velocity of a satellite is

$7.613\ kms^{-1}$ .

Time period of a satellite:

The time taken by a satellite to complete one revolution around the Earth is called the time period (

$T$ ).

$\begin{array}{l} Time period = \frac{Distance covered}{Orbital velocity} \\ T = \frac{2 π r}{v} \end{array}$

On substituting the values of

$v$ , we get

$T = \frac{2 π (R + h)}{\sqrt{\frac{GM}{(R + h)}}}$

Example:

Find the orbital period of the satellite at an orbital height of

$500\ km$ .

$h$

$=$

$500\ km$

$R$

$=$

$6731\ km$

$v$

$=$

$7.613 kms{-1}$

The formula is given as

$T = \frac{2 π (R + h)}{\sqrt{\frac{GM}{(R + h)}}}$

On substituting the given values, we get

$\begin{array}{l} T = 2 \times \frac{22}{7} \times \frac{(6731 + 500)}{7. 613} \\ = 5.6677 \times 10^{3} s \\ = 5667.7 s \\ T \approx 95 \min \end{array}$

Thus, the orbital period of the satellite is

$95\ minutes$ .

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6. Calculations of orbital velocity

Theory: