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A satellite moves on an orbit of radius $44400 km$ around the earth assume only due to earth force is acting then what will be its acceleration (take weight of earth $6 \times 10^{24} kg$ ).

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0.3 m / s^{2}

0.2 m / s^{2}

0.4 m / s^{2}

0.5 m / s^{2}

answer is B.

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Detailed Solution

A satellite moves on an orbit of radius

44400 km

around the earth. The acceleration of the satellite, as the only earth force acting, is

0.20301 m / s^{2}

.
Given,
Radius of the orbit of the satellite,

r = R + h

\Rightarrow = 44400 km

\Rightarrow = 444 \times 10^{5} m

Mass of the earth,

M_{e} = 6 \times 10^{24} kg

Universal gravitational constant,

G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

Now,
the acceleration of the satellite,

a = (G \times M_{e}) / r^{2}

\Rightarrow = {(6.67 \times 10^{- 11}) \times (6 \times 10^{24})} / {(444 \times 10^{5})}^{2}

\Rightarrow = (40.02 \times 10^{13}) / (197136 \times 10^{10})

\Rightarrow = 0.20301 m / s^{2}

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