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

Question

ol3i4c5s4hr · Accepted Answer

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

Dabanka4v · Accepted Answer

Explanation:
The period of the satellite can be determined by;
$T = \sqrt{\frac{4 π^{2} r^{2}}{G M}}$
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: $r = 6 \times (6.38 \times 10^{6}) = 38.28 \times 10^{6} m, M = 5.98 \times 10^{24} k g, G = 6.673 \times 10^{- 11} N \frac{m^{2}}{k} g^{2}$
Therefore, $T = \sqrt{\frac{4 \cdot {(\frac{22}{7})}^{2} {(38.28 \cdot 10^{6})}^{2}}{6.673 \cdot 10^{- 11} \cdot 5.98 \cdot 10^{24}}}$
$= \sqrt{\frac{5.79 \cdot 10^{16}}{3.99 \cdot 10^{14}}}$
$= \sqrt{145.113}$
$= 12.0463$
$T = 12.05 s$
The period of the satellite is 12.05s

A satellite circles the earth in an orbit whose radius

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