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Gravitation physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eight

A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let r be the distance of the body from the centre of the star and let its linear velocity be v, angular velocity ω, kinetic energy K, gravitational potential energy U, total energy E and angular momentum l. As the radius r of the orbit increases, determine which of the above quantities increase and which ones decrease.

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Type Questions as classified in NCERT Exemplar

Explanation- when a body og mass m is revolving around a star of mass M.

Linear velocity of the body v= $\sqrt[]{\frac{G M}{r}}$ so when r increases then v decreases.

Angular velocity of the body w = 2 $π / T$

According to kepler’s law T² $\propto$ r³

So T= kr^3/2

So w= $\frac{2 π}{k r^{3 / 2}}$ so when r increases, w decreases.

Kinetic energy of the body K= 1/2mv²=1/2m ( $\frac{G M}{r}$ ) so when we increase r, KE decreases.

Gravitational potential energy of the body

U=-GMm/r

So when we increase r, PE becomes less negative

Total energy of the body E= KE+PE= $\frac{G M m}{2 r} - \frac{G M m}{r} = - \frac{G M m}{2 r}$

When r increases total energy becomes less negative . i.e increases.

Angul

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This is a Long Type Questions as classified in NCERT Exemplar

Explanation- when a body og mass m is revolving around a star of mass M.

Linear velocity of the body v= $\sqrt[]{\frac{G M}{r}}$ so when r increases then v decreases.

Angular velocity of the body w = 2 $π / T$

According to kepler’s law T² $\propto$ r³

So T= kr^3/2

So w= $\frac{2 π}{k r^{3 / 2}}$ so when r increases, w decreases.

Kinetic energy of the body K= 1/2mv²=1/2m ( $\frac{G M}{r}$ ) so when we increase r, KE decreases.

Gravitational potential energy of the body

U=-GMm/r

So when we increase r, PE becomes less negative

Total energy of the body E= KE+PE= $\frac{G M m}{2 r} - \frac{G M m}{r} = - \frac{G M m}{2 r}$

When r increases total energy becomes less negative . i.e increases.

Angular momentum of the body L =mvr =mr $\sqrt[]{\frac{G M}{r}} =$ m $\sqrt[]{G M r}$

L= $\sqrt[]{r}$

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This is a Long Type Questions as classified in NCERT ExemplarExplanation- when a body og mass m is revolving around a star of mass M.<picture><img src="https://images.shiksha.com/mediadata/images/articles/1745474520phpM4cPou.jpeg" alt="" width="646" height="363"></picture>Linear velocity of the body  v= <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math>  so when r increases then v decreases.Angular velocity of the body w = 2 <math> <mi>π</mi> <mo>/</mo> <mi>T</mi> </math> According to kepler’s law T2 <math> <mo>∝</mo> </math> r3 So T= kr3/2So w= <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mi> </mi> </math> so when r increases, w decreases.Kinetic energy of the body K= 1/2mv2=1/2m ( <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> )  so when we increase r, KE decreases.Gravitational potential energy of the bodyU=-GMm/rSo when we increase r, PE becomes less negativeTotal energy of the body E= KE+PE= <math> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </mrow> </mfrac> </math> When r increases total energy becomes less negative . i.e increases.Angular momentum of the body L =mvr =mr <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>G</mi> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> </math> m <math> <mroot> <mrow> <mrow> <mi>G</mi> <mi>M</mi> <mi>r</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> L= <math> <mroot> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math>

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