Gravitation

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b

Explanation- as observed from the sun two types of forces are acting on the moon one is due to gravitational attraction between the sun and the moon and the other is due to gravitational attraction between the earth and the moon hence total force is not normal.

This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer-c

Explanation- As the total energy of the earth satellite bounded system is negative . due to the viscous force acting on the satellite, energy decreases continuously and radius also decreases.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a

Explanation- As the earth is revolving around the sun in a circular motion due to gravitational attraction. The force of attraction will be of radial nature and angle between them is zero so torque is also zero.

Answer-c

Explanation-The gravitational force between sun and earth follows inverse square law. Due to relative motion between earth and mercury, the orbit of mercury observed from the earth will not be approximately circular since the major gravitational force on mercury is due to the sun.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- We treated that the mass of earth is at centre . in this case a=g=0 at centre but if earth is not uniform then value of g is different at different point.

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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

This is a  Short Answer Type Questions as classified in NCERT Exemplar

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 = $\frac{GMm}{2R}-\frac{GMm}{R}$ = - $\frac{GMm}{2R}$

This is a  Short Answer Type Questions as classified in NCERT Exemplar

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.