Question

What linear speed must an earth satellite have to be on a circular orbit of altitude $160{\text{ km}}$ above earth’s surface? Also find the period of revolution?

Answer 1

We will find the gravitational force acting on the satellite due to earth. Also when the satellite moves in orbit then it will have some acceleration. By applying Newton’s second law of motion we can find the linear speed of the earth satellite at a distance of $160{\text{ km}}$ above earth’s surface. Period of revolution can be found by dividing the distance covered by the satellite in one revolution by its linear speed.

Formula Used:
$(i){\text{ F = }}\dfrac{{G{M_e}{m_s}}}{{{r^2}}}$
Where, $G{\text{ = }}$ Gravitational constant, ${{\text{M}}_e}{\text{ = }}$ Mass of earth, ${m_s}{\text{ = }}$ Mass of satellite and $r{\text{ = }}$ Orbital radius.
$(ii){\text{ F = }}\dfrac{{{m_s}{v^2}}}{r}$
Where, $v{\text{ = }}$ Speed of satellite, ${m_s}{\text{ = }}$ Mass of satellite and $r{\text{ = }}$ Orbital radius.

Complete step by step answer:
According to Newton’s Law of gravitation the gravitational force between earth and satellite will be equal to:
${\text{ F = }}\dfrac{{G{M_e}{m_s}}}{{{r^2}}}$ ____________$(1)$

Now when satellite revolves around the around then it experience an centripetal force due to its motion and this force will be equal to:
${\text{F = }}\dfrac{{{m_s}{v^2}}}{r}$ ____________$(2)$
On comparing equation $(1)$ and $(2)$ we can write that:
$\dfrac{{G{M_e}{m_s}}}{{{r^2}}}{\text{ = }}\dfrac{{{m_s}{v^2}}}{r}$
On cancelling the common terms on both side the equation for speed will be,
$v{\text{ = }}\sqrt {\dfrac{{G{M_e}}}{r}}$
Orbital radius $\left( r \right)$ is the sum of the radius of earth and height of the satellite above the surface. We know that radius of earth is $6.36{\text{ }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}$ and height of satellite above earth surface is given as $160{\text{ km}}$, thus orbital radius will be:
Orbital radius $= {\text{ }}6.36{\text{ }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m + 160 }} \times {\text{ 1}}{{\text{0}}^3}{\text{ m}}$
Orbital radius $= {\text{ 6}}{\text{.53 }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}$

Also we know that gravitational constant $(G)$ is equal to $6.67{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ }}{{\text{m}}^3}{\text{ }}{{\text{s}}^{ - 2}}{\text{ k}}{{\text{g}}^{ - 1}}$ and mass of earth $\left( {{M_e}} \right){\text{ = 6 }} \times {\text{ 1}}{{\text{0}}^{24}}{\text{ kg}}$. On substituting the values we get the speed of satellite as:
$v{\text{ = }}\sqrt {\dfrac{{G{M_e}}}{r}}$
$\Rightarrow v{\text{ = }}\sqrt {\dfrac{{6.67{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 11}}{\text{ }}{{\text{m}}^3}{\text{ }}{{\text{s}}^{ - 2}}{\text{ k}}{{\text{g}}^{ - 1}}{\text{ }} \times {\text{ 6 }} \times {\text{ 1}}{{\text{0}}^{24}}{\text{ kg}}}}{{{\text{6}}{\text{.53 }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}}}}$

On solving the above we would get approximate value for velocity as:
$v{\text{ = 7}}{\text{.82 }} \times {\text{ 1}}{{\text{0}}^3}{\text{ m }}{{\text{s}}^{ - 1}}$
The period of revolution can be calculated as the ratio of distance travelled by the satellite and speed of the satellite.
Period of revolution $= {\text{ }}\dfrac{{{\text{Distance Travelled}}}}{{Speed}}$
It can be deduced as,
Period of revolution $= {\text{ }}\dfrac{{2\pi r}}{v}$

On substituting the values we get the time period of revolution as,
Period of revolution $= {\text{ }}\dfrac{{2{\text{ }} \times {\text{ 3}}{\text{.14 }} \times {\text{ 6}}{\text{.53 }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}}}{{7.82{\text{ }} \times {\text{ 1}}{{\text{0}}^3}{\text{ m }}{{\text{s}}^{ - 1}}}}$
Period of revolution $= {\text{ }}\dfrac{{{\text{41 }} \times {\text{ 1}}{{\text{0}}^6}{\text{ m}}}}{{7.82{\text{ }} \times {\text{ 1}}{{\text{0}}^3}{\text{ m }}{{\text{s}}^{ - 1}}}}$
Period of revolution $= {\text{ 5}}{\text{.25 }} \times {\text{ 1}}{{\text{0}}^3}{\text{ s}}$
In minutes it can be represented as,
Period of revolution $= {\text{ 87}}{\text{.5}}$ Minutes

Hence, the linear speed is ${\text{7}}{\text{.82 }} \times {\text{ 1}}{{\text{0}}^3}{\text{ m }}{{\text{s}}^{ - 1}}$ and period of revolution is ${\text{ 87}}{\text{.5}}$ minutes.

Note: If the mass of earth is not given in question then we can take the approximate mass of earth. The centripetal force is due to the orbital motion of the satellite while revolving around the earth. The orbital radius is measured from the centre of earth, that’s why we have to add the radius of earth to the height of the satellite above earth’s surface.