Gravitational Constant Class 11th Physics Physics Gravitation

Four particle each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mroot> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>M</mi> </mrow> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mroot> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>M</mi> </mrow> <mrow> <mi>R</mi> </mrow> </mfrac> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mroot> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>M</mi> </mrow> <mrow> <mi>R</mi> </mrow> </mfrac> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>M</mi> </mrow> <mrow> <mi>R</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

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Alok Kumar Singh | Contributor-Level 10

7 months ago

Correct Option - 3

Detailed Solution:

For circular motion F_{net $= \frac{M V^{2}}{R}$}

_{$\sqrt[]{2 F} + F' = \frac{M V^{2}}{R}$}

$\sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} R)}^{2}} + \frac{G M M}{{(2 R)}^{2}} = \frac{M V^{2}}{R}$

$\frac{G M}{R} [\frac{1}{\sqrt[]{2}} + \frac{1}{4}] = V^{2}$

$V = \frac{1}{2} \sqrt[]{\frac{G M}{R} (2 \sqrt[]{2} + 1)}$

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The satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of nearer satellite is 2 × 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{π}{x} r a d h^{- 1}$ where x is__________.

Alok Kumar Singh

For A satellite T₁ = 1hour

$S o ω_{1} = 2 π r e d / h o u r$

for B satellite T₂ = 8 hour

given R₁ = 2 × 10³ Km

Relative $ω = \frac{V_{1} - V_{2}}{R_{2} - R_{1}} = \frac{2 π \times 1 0^{3}}{6 \times 1 0^{3}}$

$\Rightarrow \frac{π}{3}$ rad/hour

x = 3

A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100kg and radius 50m. If the gravitational potential at a point, 25m from the centre is V kg/m. The value of V is:

Alok Kumar Singh

$V_{A} = - \frac{G M_{1}}{r} - \frac{G M_{2}}{R} = - \frac{50 G}{25} - \frac{100 G}{50} = - 4 G$

Two satellites S₁ and S₂ are revolving in circular orbits around a planet with radius R₁ = 3200km and R₂ = 800km respectively. The ratio of speed of satellite S₁ to the speed of satellite S₂ in their respective orbits would be $\frac{1}{x}$  where x =…………..

Alok Kumar Singh

M₁ ->Mass of satellite (1)

M₂ -> Mass of satellite (2)

M_P -> Mass of planet

Now NLM (2) on S₁

_{$\frac{G M_{1} M_{P}}{R_{1}^{2}} = \frac{m_{1} v_{1}^{2}}{R_{1}}$}

Similarly v₂ = $\sqrt[]{\frac{G M_{P}}{R_{2}}}$

$\frac{v_{1}}{v_{2}} = \sqrt[]{\frac{R_{2}}{R_{1}}} = \sqrt[]{\frac{800}{3200}} = \frac{1}{2} = \frac{1}{x}$

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