Physics NCERT Exemplar Solutions Class 11th Chapter Eight

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Explanation- Gravitational force at h height F = $\frac{GMmh}{{{(r}^2+h^2)}^{3/2}}$

When mass is displaced upto distance 2h then

F'= $\frac{GMm2h}{{{[r}^2+{\left(2h\right)]}^2]}^{\frac{3}{2}}}$

= $\frac{2GMmh}{{\left(r^2+4h^2\right)}^{\frac{3}{2}}}$

When h = r then F= $\frac{GMmh}{[{r^2+{\left(r\right)}^2]}^{3/2}}$

So F = $\frac{GMm}{2\sqrt[]{2r^2}}$

F'= $\frac{2GMmr}{{\left(r^2+4r^2\right)}^{\frac{3}{2}}}=\frac{2GMm}{5\sqrt[]{5r^2}}$

$\frac{F^{\text{'}}}{F}=\frac{4\sqrt[]{2}}{5\sqrt[]{5}}$

F' = $\frac{4\sqrt[]{2}}{5\sqrt[]{5}}F$

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Explanation-potential energy of the objects at the surface of the earth is =- $\frac{GMm}{R}$

PE of the object at a height equal to radius of earth = $\frac{GMm}{2R}$

Gain in potential energy = $\frac{GMm}{2R}$ - (- $\frac{GMm}{R}$ )= $\frac{GMm}{2R}$

= $\frac{g{R}^{2}m}{2R}=\frac{1}{2}mgR$ (GM=gR²)

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Explanation-The trajectory of a particle under gravitational force of the earth will be in conic section with the centre of the earth as a focus. Only c meets this requirements

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Explanation- Orbital speed of the satellite v_o= $\sqrt[]{\frac{GM}{R}}$

KE of the satellite of mass m, E_k = 1/2mv_o²= Gmm/2R              

So kinetic energy is inversely proportional to distance.

b)potential energy of a satellite E_p=-GMm/R

so it is also inversely proportional to R

c)total energy of the satellite E=E_k+E_p = $\frac{GMm}{2R}-\frac{GMm}{R}$ = - $\frac{GMm}{2R}$

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Explanation- Let M be the mass and R be the radius of sphere . an object of mass m be placed at the mid point P of the line joining their centres

Force F₁=F₂=GMm/ (5R)²

If we displace it through x distance

New force towards sphere A, F₁'= GMm/ (5R-x)²

Sphere B F₂' = GMm/ (5R+x)²

As F₁'>F₂' therefore a resultant force (F₁'-F₂') acts on the objects towards sphere A therefore objects start to move towards A hence equilibrium is unstable.

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Explanation-Every day the earth advances in the orbit by approximately 1⁰. Then it will have to rotate by 361⁰ to have the sun at zenith point again. So 361⁰ corresponds to 24h

So 1^o corresponds to 24/361= 0.066h=3.99 min =4min

Hence distant stars would rise 4 min early every successive day.

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Explanation- Angle between the equatorial plane and orbital plane of a polar satellite is 90⁰ and angle between equatorial plane and orbital plane of a geostationary satellite is 0⁰.

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Explanation- Phelion is the location of the earth where it is the greatest distance from the sun and perihelion is the location of the earth where it is the nearest to the sun

The areal velocity (1/2 rv) of the earth around the sun is constant . so speed of earth is more at perihelion.

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Explanation- As density of the shell is constant

So mass of shell = density (volume)

= $\rho \left(\frac{4}{3}\pi R^3\right)=M$

If we consider shell is uniform so F=gravitational force =GMm/r²

F=0 for r

= GM/r² for r>R

Variation of f verses r

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Explanation- Inside a small spaceship orbiting around the earth, the value of acceleration due to gravity can be calculated as constant and hence astronaut feels weightlessness.

If the space station orbiting around the earth has a large size such that variation in g matters in that case astronaut inside the spaceship will experience gravitational force and hence can detect gravity.