The time period of satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is:

Option 1 - 12 T

Option 2 - 9 T

Option 3 - 27 T

Option 4 - 3 T

6 Views|Posted 5 months ago

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Answered by

Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 3

Detailed Solution:

T² ∝ R³
⇒ (T? /T? )² = (R? /R? )³ ⇒ T? = 9³/² T = 27T

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Similar Questions for you

Two artificial satellites of masses $m_{1}$ and $m_{2}$ are moving with speeds $v_{1}$ and $v_{2}$ in orbits of radii $r_{1}$ and $r_{2}$ , respectively. If r₁>r₂ then which of the following statements is true

Raj Pandey

According to kepler’s third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4 r_{B}^{3}$

A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02R. The percentage difference in the time periods of the two satellites is:

Vishal Baghel

Time period of Satellite = T = $\frac{2 π R}{v} = {(\frac{4 π^{2} R^{3}}{G M})}^{\frac{1}{2}}$

$\Rightarrow T α R^{\frac{3}{2}} \Rightarrow \frac{T_{2}}{T_{1}} = \frac{{(1.02 R)}^{\frac{3}{2}}}{R^{\frac{3}{2}}} = {(1.02)}^{\frac{3}{2}} = {(1 + 0.02)}^{\frac{3}{2}}$

The percentage difference in the time periods of the two satellites = $\frac{Δ T}{T_{1}}$ × 100

= (0.03) × 100 = 3%

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