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