Question 73: Gravitation

Question

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

Accepted Answer

$$ T \propto R^{\frac{3}{2}} $$
For a satellite in a circular orbit with radius $$ R $$, the time period is given by
$$ T = 2\pi \sqrt{\frac{R^3}{GM}}. $$
If the radius is increased to $$ 1.03R $$, the new time period $$ T' $$ becomes
$$ T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T 	imes (1.03)^{\frac{3}{2}}. $$
For a small change, we can approximate
$$ (1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} 	imes 0.03 = 1 + 0.045 = 1.045. $$
This means the new time period is approximately 4.5% larger than the original time period.
Thus, the time period of the second satellite is larger by approximately 4.5%.

Answer

$3 \%$

Answer

$2.5 \%$

Answer

$4.5 \%$

Answer

$9 \%$

Gravitation

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

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