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2021-12-20

A satellite circles the earth in an orbit whose radius is twice the earth s

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Step 1
Concept:
As we know, If an object is moving in a circle of radius r with constant speed v, then the motion is said to be a uniform circular motion. It has a radial acceleration ${a}_{R}$ that is directed radially toward the center of the circle which is called radial acceleration which is given by
${a}_{R}=\frac{{v}^{2}}{r}$
The velocity vector and the acceleration vector ${a}_{R}$ are continually changing in direction, but are perpendicular to each other at each moment. When the speed of circular motion is not constant, the acceleration has two components, tangential as well as centipental. As we know Newtons

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Explanation:
The period of the satellite can be determined by;
$T=\sqrt{\frac{4{\pi }^{2}{r}^{2}}{GM}}$
where: T is the period, r is the radius of the orbit, G is the gravitation constant and M is the mass of the Earth.
Given that: $r=6×\left(6.38×{10}^{6}\right)=38.28×{10}^{6}m,M=5.98×{10}^{24}kg,G=6.673×{10}^{-11}N\frac{{m}^{2}}{k}{g}^{2}$
Therefore, $T=\sqrt{\frac{4\cdot {\left(\frac{22}{7}\right)}^{2}{\left(38.28\cdot {10}^{6}\right)}^{2}}{6.673\cdot {10}^{-11}\cdot 5.98\cdot {10}^{24}}}$
$=\sqrt{\frac{5.79\cdot {10}^{16}}{3.99\cdot {10}^{14}}}$
$=\sqrt{145.113}$
$=12.0463$
$T=12.05s$
The period of the satellite is 12.05s

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