Question

Supppose the orbit of a satellite is exactly 35,780 km above the earth 's surface. Determine the tangential velocity of the satellite

Answer 1

Here's the answer

We know the following values =>

$\bf{G = 6.673 \times 10 {}^{ - 11} Nm^2 / kg ^2}$

Mass of earth = $\bf { 6 \times 10 ^24 km }$

Radius of the earth = $\bf { 6400 km = 6400000 m }$

h = height of the satellite above the Earth's suface 35780 km = 35780000 m

To find => v = ?

Solution =>

We know the formula ..

$\bf{v = \sqrt{ \frac{GM}{ R \: + h} } }$

$\sf{v = \sqrt{ \frac{6.67 \times 10 {}^{ - 11 } \times 6 \times 10 {}^{24} }{357800000 + 6400000}} }$

$\sf{v = \sqrt{ \frac{40.02 \times 10 {}^{13} }{42180000}} }$

$\sf{v = \sqrt{ \frac{40.02 \times 10 {}^{13} }{42.18 \times 10 {}^{6}} } }$

$\sf{v = \sqrt{ \frac{40 \times 10 {}^{7} }{42}}}$

$\sf{v = \sqrt{ \frac{0.95 \times 10 {}^{7} }{9.5 \times 10 {}^{6}} } }$

$\sf{v = 3.08 \times 10 {}^{3} m/s}$

$\bf {\therefore v = 3.08 \: km/s}$

Final answer =>

${ \boxed {\boxed{\sf{Tangential \: velocity \: will \: be \: 3.08 km/s}}}}$