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.