  LEARNATHON
III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

Theory:

Formula of orbital velocity:
The orbital velocity is calculated using the formula,

$v\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{\mathit{GM}}{\left(R+h\right)}}$

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

$v\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{\mathit{GM}}{\left(R+h\right)}}$

On substituting the given values, we get

$\begin{array}{l}v\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{6.673×{10}^{-11}×5.972×{10}^{24}\phantom{\rule{0.147em}{0ex}}}{\left(6731000+500000\right)}}\\ \\ =\phantom{\rule{0.147em}{0ex}}\sqrt{\frac{39.851156×{10}^{13}\phantom{\rule{0.147em}{0ex}}}{\left(7231000\right)}}\\ \\ =7613\phantom{\rule{0.147em}{0ex}}m{s}^{-1}\\ \\ =7.613\phantom{\rule{0.147em}{0ex}}\mathit{km}{s}^{-1}\end{array}$

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 ($$T$$).

On substituting the values of $$v$$, we get

$T\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{2\mathrm{\pi }\left(R+h\right)}{\sqrt{\frac{\mathit{GM}}{\left(R+h\right)}}}$

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

$T\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{2\mathrm{\pi }\left(R+h\right)}{\sqrt{\frac{\mathit{GM}}{\left(R+h\right)}}}$

On substituting the given values, we get

$\begin{array}{l}T\phantom{\rule{0.147em}{0ex}}=2×\frac{22}{7}\phantom{\rule{0.147em}{0ex}}×\frac{\left(6731+500\right)}{7.613}\\ \\ =5.6677×{10}^{3}\phantom{\rule{0.147em}{0ex}}s\\ \\ =5667.7\phantom{\rule{0.147em}{0ex}}s\\ \\ T\approx 95\phantom{\rule{0.147em}{0ex}}\mathit{min}\end{array}$

Thus, the orbital period of the satellite is $$95\ minutes$$.