Question

The time period of a satellite orbiting the Earth is 3 h. If the separation between satellite and centre of Earth is increased to 4 times the previous value, then the new time period will become
पृथ्वी का घूर्णन कर रहे एक उपग्रह का आवर्तकाल 3 h है। यदि उपग्रह तथा पृथ्वी के केन्द्र के मध्य दूरी को पूर्व मान के 4 गुने तक बढ़ाया जाता है, तब नवीन आवर्तकाल हो जाएगा
A) 36 h
B) 24 h
C) 16 h
D) 8 h
_________...
ANTHE 2020 — Physics - IXth. ​

Answer 1

Explanation:

Let T1 = 3h

T2 = x

Let the radius = R

New radius = 4R

Using, Kepler's Laws

${ (\frac{t1}{t2} )}^{2} = { (\frac{r1}{r2} )}^{3}$

${ (\frac{3}{x} )}^{2} = { (\frac{r}{4r} )}^{3}$

$\frac{9}{ {x}^{2} } = \frac{1}{64}$

${x}^{2} = 576$

$x = 24 \: h$

Therefore , b) option is correct.

I hope it helps you. If you have any doubts, then don't hesitate to ask.

Answer 2

Given :

• T₁ = 3 h
• R₁ = R
• R₂ = 4R₁ = 4R

To find :

• New time period (T₂)

Method :

Kepler's Third Law / Harmonic Law / Law of period :

The square of the period of revolution of the planet around the sun is directly proportional to the cube of the semi-major axis of the elliptical orbit.

सूर्य के चारों ओर ग्रह की क्रांति की अवधि का वर्ग अण्डाकार कक्षा के अर्ध-प्रमुख अक्ष के घन के सीधे आनुपातिक है।

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

Hence,

$\implies\sf{\bigg({\dfrac{T_1}{T_2}\bigg)}^2 = {\bigg(\dfrac{R_1}{R_2}\bigg ) }^3}$

$\implies\sf{\bigg({\dfrac{T_1}{T_2}\bigg)}^2 = {\bigg(\dfrac{R}{4R}\bigg ) }^3}$

$\implies\sf{\bigg({\dfrac{T_1}{T_2}\bigg)}^2 = \dfrac{1}{64}}$

$\implies\sf{\dfrac{T_1}{T_2} = \sqrt{ \dfrac{1}{64}}}$

$\implies\sf{\dfrac{T_1}{T_2} = \dfrac{1}{8}}$

$\implies\sf{T_2 = T_1 \times 8}$

$\implies\sf{T_2 = 3 \times 8}$

$\implies\boxed{\sf{T_2 = 24 h }}$

The new time period will be 24h.

Hence,

The answer is option B.