NCERT Solutions: Class 11 Physics Chapter 7 - Gravitation

Unlike electric fields in conductors, gravity cannot be shielded by surrounding a body with matter. In a small orbiting spaceship, astronaut and ship are in free fall together, so gravity is not locally felt. In a large station, the gravitational field is slightly different at different parts of the station, so tidal effects can reveal it. The sun's total pull on earth is larger, but tidal effect depends on the field gradient, proportional to $M/r^3$, and the moon's much smaller distance makes its tidal effect stronger.

Outside the earth, $g=GM/r^2$, so it decreases with altitude. Inside a uniform earth, $g\propto r$, so it decreases as depth increases. Acceleration due to gravity is independent of the falling body's mass. The exact gravitational potential-energy difference uses the inverse-radius formula; $mg(r_2-r_1)$ is only a near-surface approximation.

By Kepler's third law, $T^2\propto R^3$. If the planet goes around twice as fast, its period is $T_E/2$. Hence $R/R_E=(T/T_E)^{2/3}=(1/2)^{2/3}=0.63$.

For a satellite in circular orbit, $M=4\pi^2r^3/(GT^2)$. Here $T=1.769\times86400=1.528\times10^5\,\text{s}$ and $r=4.22\times10^8\,\text{m}$. Thus $M_J=\frac{4\pi^2(4.22\times10^8)^3}{(6.67\times10^{-11})(1.528\times10^5)^2}=1.90\times10^{27}\,\text{kg}$. Compared with $M_S\approx2.0\times10^{30}\,\text{kg}$, this is $9.5\times10^{-4}M_S$.

Treat the galactic mass as $M=2.5\times10^{11}(2.0\times10^{30})=5.0\times10^{41}\,\text{kg}$. The radius is $r=50000\,\text{ly}=50000(9.46\times10^{15})=4.73\times10^{20}\,\text{m}$. Using $T=2\pi\sqrt{r^3/(GM)}$, $T=1.12\times10^{16}\,\text{s}=3.55\times10^8\,\text{years}$.

For a circular orbit, $K=GMm/(2r)$ and $U=-GMm/r$, so total energy $E=K+U=-GMm/(2r)=-K=U/2$. An orbiting satellite already has kinetic energy, so the extra energy needed to escape is $GMm/(2r)$, less than the $GMm/r$ needed for a stationary object at the same radius.

Escape speed from distance $r$ is $v_e=\sqrt{2GM/r}$. The body's mass cancels out, and direction does not enter the energy condition. For a spherical earth it depends only on distance from the earth's centre, so height matters, while different locations at the same height have the same ideal escape speed.

The sun's gravitational force is central, so torque about the sun is zero and angular momentum is conserved. In an elliptical orbit the comet's distance from the sun changes, so speed, angular speed, kinetic energy and potential energy change. Since gravity is conservative and mass loss is neglected, total mechanical energy remains constant.

In weightlessness, body fluids are not pulled toward the feet as on earth, so the face may swell and headache can occur. The absence of the usual up-down cue also causes orientational problems. Swollen feet are less likely because fluid does not pool in the feet.

1. i. a
2. ii. b
3. iii. c
4. iv. 0

At the centre of the hemispherical shell, horizontal components from symmetric mass elements cancel. The remaining gravitational intensity is along the axis toward the shell material, which is arrow c in Fig. 7.11.

Let the point be at distance $x$ from earth's centre. If the earth-sun distance is $R$, set gravitational pulls equal: $\frac{GM_E}{x^2}=\frac{GM_S}{(R-x)^2}$. Thus $x=\frac{R}{1+\sqrt{M_S/M_E}}$. With $R=1.5\times10^{11}\,\text{m}$, $M_S=2\times10^{30}\,\text{kg}$ and $M_E=6\times10^{24}\,\text{kg}$, $x=2.59\times10^8\,\text{m}$.

For earth's nearly circular orbit, $GM_S/r^2=4\pi^2r/T^2$, so $M_S=4\pi^2r^3/(GT^2)$. With $r=1.5\times10^8\,\text{km}=1.5\times10^{11}\,\text{m}$ and $T=1\,\text{year}=3.156\times10^7\,\text{s}$, the mass comes out close to $2.0\times10^{30}\,\text{kg}$.

By Kepler's third law, $(T_S/T_E)^2=(R_S/R_E)^3$. Thus $R_S=R_E(29.5)^{2/3}=1.50\times10^8(29.5)^{2/3}=1.43\times10^9\,\text{km}$.

At height $h=R/2$, distance from earth's centre is $r=3R/2$. Gravitational force varies as $1/r^2$, so $F=63\left(\frac{R}{3R/2}\right)^2=63\times\frac{4}{9}=28\,\text{N}$.

Inside a uniform spherical earth, $g$ is proportional to distance $r$ from the centre. Halfway down to the centre means $r=R/2$, so the weight is half the surface weight: $250/2=125\,\text{N}$.

Use conservation of energy per unit mass. At launch, $E=\frac12v^2-GM/R$. At the highest point, speed is zero and $E=-GM/r_{max}$. Hence $\frac{1}{r_{max}}=\frac{1}{R}-\frac{v^2}{2GM}$. With $v=5000\,\text{m s}^{-1}$, $R=6.4\times10^6\,\text{m}$ and $GM=6.67\times10^{-11}\times6.0\times10^{24}$, $r_{max}=8.0\times10^6\,\text{m}$ from earth's centre. The height is $r_{max}-R=1.6\times10^6\,\text{m}$.

At infinity, $\frac12mv_{\infty}^2=\frac12m(v^2-v_e^2)$. With $v=3v_e$, $v_{\infty}=\sqrt{9v_e^2-v_e^2}=\sqrt8\,v_e=2.828(11.2)=31.7\,\text{km s}^{-1}$.

The orbital radius is $r=6.4\times10^6+4.0\times10^5=6.8\times10^6\,\text{m}$. A circular satellite has total energy $E=-GMm/(2r)$. To escape with zero speed at infinity, the required energy is $GMm/(2r)=\frac{(6.67\times10^{-11})(6.0\times10^{24})(200)}{2(6.8\times10^6)}=5.9\times10^9\,\text{J}$.

Initial separation is $r_i=10^9\,\text{km}=10^{12}\,\text{m}$. They touch when centre separation is $r_f=2(10^4\,\text{km})=2\times10^7\,\text{m}$. By energy conservation, total kinetic energy at collision is $Mv^2$ for two equal masses each moving with speed $v$. Thus $Mv^2=GM^2(1/r_f-1/r_i)$, so $v=\sqrt{GM(1/r_f-1/r_i)}=2.6\times10^6\,\text{m s}^{-1}$.

The midpoint is $0.5\,\text{m}$ from each sphere's centre. The gravitational fields due to the two equal spheres are equal and opposite, so net force is zero. Potential adds algebraically: $V=-\frac{GM}{0.5}-\frac{GM}{0.5}=-\frac{2G(100)}{0.5}=-2.67\times10^{-8}\,\text{J kg}^{-1}$. A small displacement along the line makes the nearer sphere pull more strongly, so the object moves farther away from the midpoint; the equilibrium is unstable.