Physics, asked by Ananyasingh8397, 11 months ago

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

Answers

Answered by CarliReifsteck
79

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

Answered by swati1212
17

Answer:

please refer to the attachment

Attachments:
Similar questions