A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01)R . The period of the second satellite is larger than that of the first one by approximately

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 (1.01)R . The period of the second satellite is larger than that of the first one by approximately (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

In the problem orbital radius is increased by

1%

Time period of satellite

T \propto r^{3/2}

\Rightarrow \frac{Δ T}{T} = \frac{3}{2} \times \frac{Δ r}{r}

Percentage change in the time period

= \frac{3}{2}

[

%

change in orbital radius]

= \frac{3}{2} (1%) = 1.5%

Q. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01)R . The period of the second satellite is larger than that of the first one by approximately

In the problem orbital radius is increased by 1% Time period of satellite T∝r3/2 ⇒TΔT​=23​×rΔr​ Percentage change in the time period =23​ [ % change in orbital radius] =23​(1%)=1.5%

Q. A satellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $(1.01) R$ . The period of the second satellite is larger than that of the first one by approximately

In the problem orbital radius is increased by $1%$
Time period of satellite $T \propto r^{3/2}$
$\Rightarrow \frac{Δ T}{T} = \frac{3}{2} \times \frac{Δ r}{r}$
Percentage change in the time period
$= \frac{3}{2}$ [ $%$ change in orbital radius]
$= \frac{3}{2} (1%) = 1.5%$