Question

A remote sensing satellite of earth revolves in a circular orbit at the height of 0.25 in 10 power 6 m above the surface of earth if radius is 6.38 mm 6 m and g equal to 9.8 metre per second square then the orbital speed of the satellite is

Answer 1

Answer:

The orbital speed of the satellite is 7.76 km/s

Explanation:

Given that,

Height $h=0.25\times10^{6}\ m$

Radius $r= 6.38\times10^{6}\ m$

We need to calculate the orbital speed of the satellite

Using centripetal force and gravitational force

$\dfrac{mv^2}{R_{0}^2}=\dfrac{GmM}{R_{0}^2}$

$v=\sqrt{\dfrac{GM}{R_{0}}}$...(I)

We know that,

The formula of gravity

$g=\dfrac{GM}{R_{e}^{2}}$

$M=\dfrac{gR_{e}^2}{G}$

Put the value of g in equation (I)

$v=\sqrt{\dfrac{G\times\dfrac{gR_{e}^2}{G}}{(R_{e}+h)}}$

$v=\sqrt{\dfrac{gR_{e}^2}{(R_{e}+h)}}$

Put the value into the formula

$v=\sqrt{\dfrac{9.8\times(6.38\times10^{6})^2}{(6.38\times10^{6}+0.25\times10^{6})}}$

$v=7756.6\ m/s$

$v=7.76\ km/s$

Hence, The orbital speed of the satellite is 7.76 km/s

Answer 2

Answer:

please refer to the attachment