1.
A shell is fired vertically from the earth with speed $$\frac{{{V_{{\text{esc}}}}}}{N},$$ where $$N$$ is some number greater than one and $${{V_{{\text{esc}}}}}$$ is escape speed for the earth. Neglecting the rotation of the earth and air resistance, the maximum altitude attained by the shell will be ($${R_E}$$ is radius of the earth)
By conservation of energy
$$\eqalign{
& - \frac{{GMm}}{{{R_E}}} + \frac{1}{2}\frac{m}{{{N^2}}}\frac{{GM}}{{2{R_E}}} = - \frac{{GMm}}{H} \cr
& \Rightarrow H = \frac{{{N^2}{R_E}}}{{{N^2} - 1}} \cr
& {\text{Altitude}} = H - R = \frac{{{R_E}}}{{{N^2} - 1}} \cr} $$
2.
Imagine a new planet having the same density as that of the earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of the earth is $$g$$ and that on the surface of the new planet is $$g',$$ then
The acceleration due to gravity on the new planet can be found using the relation
$$g = \frac{{GM}}{{{R^2}}}\,......\left( {\text{i}} \right)$$
but $$M = \frac{4}{3}\pi {R^3}\rho ,\rho $$ being density.
Thus, Eq. (i) becomes
$$\eqalign{
& \therefore g = \frac{{G \times \frac{4}{3}\pi {R^3}\rho }}{{{R^2}}} = G \times \frac{4}{3}\pi R\rho \cr
& \Rightarrow g \propto R \cr
& \therefore \frac{{g'}}{g} = \frac{{R'}}{R} \cr
& \Rightarrow \frac{{g'}}{g} = \frac{{3R}}{R} = 3 \Rightarrow g' = 3g \cr} $$
3.
The speed of earth’s rotation about its axis is $$\omega .$$ Its speed is increases to $$x$$ times to make the effective acceleration due to gravity equal to zero at the equator. Then $$x$$ is :
The weight ($$= mg$$ ) of the body at the centre of the earth is zero, because the value of $$g$$ at centre is zero.
6.
A body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest
A.
at the instant just before the body hits the earth
B.
it remains constant all through
C.
at the instant just after the body is projected
D.
at the highest position of the body
Answer :
at the instant just before the body hits the earth
We know that, Power, $$P = F \cdot V = FV\cos \theta $$
So, just before hitting, $$\theta $$ is zero, power will be maximum.
7.
Kepler’s third law states that square of period of revolution $$\left( T \right)$$ of a planet around the sun, is proportional to third power of average distance $$r$$ between the sun and planet i.e. $${T^2} = K{r^3},$$ here $$K$$ is constant. If the masses of the sun and planet are $$M$$ and $$m$$ respectively, then as per Newton’s law of gravitation force of attraction between them is
$$F = \frac{{GMm}}{{{r^2}}},$$ here $$G$$ is gravitational constant. The relation between $$G$$ and $$K$$ is described as
The gravitational force of attraction between the planet and sun provide the centripetal force
i.e. $$\frac{{GMm}}{{{r^2}}} = \frac{{m{v^2}}}{r}$$
$$ \Rightarrow v = \sqrt {\frac{{GM}}{r}} $$
The time period of planet will be
$$T = \frac{{2\pi r}}{v} \Rightarrow {T^2} = \frac{{4{\pi ^2}{r^2}}}{{\frac{{GM}}{r}}} = \frac{{4{\pi ^2}{r^3}}}{{GM}}\,......\left( {\text{i}} \right)$$
Also from Kepler's third law
$${T^2} = K{r^3}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\frac{{4{\pi ^2}{r^3}}}{{GM}} = K{r^3} \Rightarrow GMK = 4{\pi ^2}$$
8.
A particle of mass $$10 \,g$$ is kept on the surface of a uniform sphere of mass $$100 \,kg$$ and radius $$10 \,cm.$$ Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $$G = 6.67 \times {10^{ - 11}}N{m^2}/k{g^2}$$ )
Acceleration due to gravity $$\left( g \right)$$ is given by
$$g = \frac{{GM}}{{{R^2}}} \Rightarrow g \propto \frac{1}{{{R^2}}}$$
As one moves from the equator to the poles, the radius of the earth decreases, hence $$g$$ increases.