A satellite is launched into a circular orbit of radius ' R ' around the earth. A second satellite is launched into an orbit of radius 1.03 R . The time period of revolution of the second satellite is larger than the first one approximately by
Solution
<p>$T \propto R^{\frac{3}{2}}$</p>
<p>For a satellite in a circular orbit with radius $R$, the time period is given by</p>
<p>$T = 2\pi \sqrt{\frac{R^3}{GM}}.$</p>
<p>If the radius is increased to $1.03R$, the new time period $T'$ becomes</p>
<p>$T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T \times (1.03)^{\frac{3}{2}}.$</p>
<p>For a small change, we can approximate</p>
<p>$(1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} \times 0.03 = 1 + 0.045 = 1.045.$</p>
<p>This means the new time period is approximately 4.5% larger than the original time period.</p>
<p>Thus, the time period of the second satellite is larger by approximately 4.5%.</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Satellites and Orbital Motion
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