Question

Two satellites move in circular orbits around the earth at distances 9000 km and 9010 km from earth's centre. If the satellite which is moving faster has a period of revolution 90 minutes. Find the difference in their revolution periods.

Answer 1

Answer:

Given:

Radius of the circular orbits are:

• 9000 km
• 9010 km

Period of revolution of faster satellite is 90 minutes.

To find:

Difference in time period of revolution.

Calculation:

According to Kepler's Law of Planetary Motion , we can say that :

${T}^{2} \propto {R}^{3}$

For the 1st satellite :

${T_{1}}^{2} \propto {R_{1}}^{3}$

$=>{90}^{2} \propto {9000}^{3}$

For 2nd satellite :

${T_{2}}^{2} \propto {R_{2}}^{3}$

$=>{T_{2}}^{2} \propto {9010}^{3}$

Dividing the proportionality :

${ \bigg(\dfrac{T_{2}}{T_{1}}\bigg)}^{2}={\bigg(\dfrac{9010}{9000}\bigg)}^{3}$

$=>{\bigg(\dfrac{T_{2}}{90}\bigg)}^{2}={\bigg(\dfrac{9010}{9000}\bigg)}^{3}$

$=>{\bigg(\dfrac{T_{2}}{90}\bigg)}^{2}=1.0033$

$=>\dfrac{T_{2}}{90}= \sqrt{1.0033}$

$=>\dfrac{T_{2}}{90}= 1.0016$

$=> T_{2} = 90.15 \:min$

So Difference in time period :

$\Delta T = 90.15 - 90 = 0.15 \:min$

So final answer :

$\boxed{\sf{\huge{\red{\bold{\Delta t = 0.15 min}}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• First satellite distance from Earth (r1) = 9000 km
• Second satellite distance from Earth (r2) = 9010 km
• Time interval of First Satellite (T1) = 90 mins

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To Find :

• Difference in time interval of both the satellite

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Solution :

Use Keppler's 3rd Law :

$\large {\boxed{\sf{T^2 \: = \: kR^3}}}$

Take case of 1st Satellite

$\implies {\sf{T_1 ^2 \: = \: k R_1 ^2}}----(1)$

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Now, take care case of 2nd Satellite

$\implies {\sf{T_2 ^2 \: = \: k R_2 ^3}} -----(2)$

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Divide equation 2 by 1

$\implies {\sf{\dfrac{T_2 ^2}{T_1 ^2} \: = \: \dfrac{k R_2 ^3}{k R_1 ^3}}} \\ \\ \implies {\sf{ \bigg( \dfrac{T_2}{T_1} \bigg) ^2 \: = \: \dfrac{R_2 ^3}{R_1 ^3}}} \\ \\ \implies {\sf{\dfrac{T_2}{T_1} \: = \: \bigg( \dfrac{R_2}{R_1} \bigg) ^{\frac{3}{2}} }} \\ \\ \implies {\sf{T_2 \: = \: T_1 \bigg( \dfrac{9010}{9000} \bigg) }} \\ \\ \implies {\sf{T_2 \: = \: 90.1500 \: mins}}$

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So, the difference in time interval is :

$\implies {\sf{Difference \: = \: T_2 \: - \: T_1}} \\ \\ \implies {\sf{Difference \: = \: 90.1500 \: - \: 90}} \\ \\ \implies {\sf{Difference \: = \: 0.15}} \\ \\ \underline{\sf{\therefore \: Difference \: in \: Period \: is \: 0.15 \: mins}}$