ncert solutions physics class 11th

Yes, a body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, $f_g$ = $\frac{GMm}{R^2}$ , where M = mass of the star = 2.5 $*2*{10}^{30}$ = 5 $*{10}^{30}$ kg

M = mass of the body, R = radius of the star = 12km = 1.2 $*{10}^{4}m$

$f_g$ = $\frac{6.67*{10}^{-11}*5*{10}^{30}*m}{{{(1.2*{10}^4)}^2}^{}}=$ 2.31 $*{10}^{12}mN$

Centrifugal force $f_c$ = mr ${\omega }^2$ where $\omega =$ angular speed = 2 $\pi \gamma$ and angular frequency $\gamma$ = 1.2 rev/s

$f_c=$ mR( ${2\pi \gamma )}^2$ = m $*$ (1.2 $*{10}^4)*2*3.14*1.2*(2*3.14*1.2)$ = 6.81 $*{10}^{5}mN$

Since $f_g>f_c$ , the body will remain stuck to the surface of the star.

Mass of the Earth, M = 6.0 $*{10}^{24}$ kg

Radius of the Earth, R = 6400 km = 6.4 $*{10}^6$ m

Height of geostationary satellite from the surface of the Earth, h = 36000 km = 3.6 $*{10}^7$ m

Gravitational potential energy due to Earth's gravity at height h:

= $\frac{-GM}{(R+h)}$

= $-\frac{6.67*{10}^{-11}*6.0*{10}^{24}}{3.6*{10}^7+6.4*{10}^6}$ = $-\frac{6.67*6.0*{10}^{13}}{4.24*{10}^7}$ = $-$ 9.44 $*{10}^6$ J/kg

Mass of each sphere, M = 100 kg

Separation between the spheres, r = 1 m

X is the midpoint between the spheres.

Gravitational force at point x will be zero. This is because gravitational force exerted by each spheres will act in opposite directions.

Gravitational potential at point x:

= $\frac{-GM}{\frac{r}{2}}$ $\frac{-GM}{\frac{r}{2}}$ = $-$ 4 $\frac{GM}{r}$ = $-\frac{4*6.67*{10}^{-11}*100}{1}$ = $-2.668*{10}^{-8}$ J/kg

Any object placed at point x will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

Escape velocity of the projectile on the Earth's surface, $v_{esc}$ = 11.2 km/s = 11.2 ${*10}^3$ m/s

Projection velocity of the projectile = 3 $v_{esc}$

Mass of the projectile = m

Velocity of the projectile far away from the Earth = $v_f$

Total energy of the projectile on the Earth = $\frac{1}{2}$ m ${v_p}^2$ $-$ $\frac{1}{2}$ m ${v_{esc}}^2$

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = $\frac{1}{2}$ m ${v_f}^2$

From the law of conservation of energy

$\frac{1}{2}$ m ${v_p}^2$ $–$ $\frac{1}{2}$ m ${v_{esc}}^2$ = $\frac{1}{2}$ m ${v_f}^2$

${v_p}^2-{v_{esc}}^2={v_f}^2$

$v_f$ = $\sqrt{{v_p}^2-{v_{esc}}^2}$ = $\sqrt{{{(3v}_{esc})}^2-{v_{esc}}^2}$ = $\sqrt{8*}$ 11.2= 31.68 km/s

Weight of a body of mass m at Earth's surface, W = mg = 250 N

Body of mass m is located at depth, d = $\frac{1}{2}{R}_e$ , where $R_e$ is radius of the Earth

Acceleration due to gravity at depth g (d) is given by : g' = (1 - $\frac{d}{R_e}$ )g = (1/2)g

Weight of the body at depth d

W' = mg' = (1/2) mg = (1/2)W = (1/2) x 250 N = 125 N

Weight of the body, W = 63 N

Acceleration due to gravity at h from the Earth's surface is given by

g' = $\frac{g}{{\left(\frac{1+h}{R_e}\right)}^2}$ where g = acceleration due to gravity on the Earth's surface,  $R_e$ = Radius of the Earth. h = $\frac{R_e}{2}$

g' = $\frac{g}{{\left(\frac{1+R_e/2}{R_e}\right)}^2}$ = $\frac{g}{{\left(\frac{1+h}{2h}\right)}^2}$ = (4/9)g

Weight of the body of mass m at a height h is given by

W' = m X g' = (4/9) mg = (4/9) x w = (4/9) x 63 N = 28 N

Distance of the Earth from the Sun, $r_e$ = 1.50 * 10⁸ km = 1.5 * 10¹¹ m

Time period of the Earth = $T_e$ , Time period of the Saturn $T_s$ = ${29.5T}_e$

Distance of Saturn from the Sun = $r_s$

From Kepler's 3rd law of planetary motion, we have T =( ${\frac{4{\pi }^2{r}^3}{GM})}^{1/2}$

For Saturn and Sun, we can write, $\frac{r_s^3}{r_e^3}$ = $\frac{T_s^2}{T_e^3}$

$r_s$ = $r_e({\frac{29.5T_e}{T_e})}^{2/3}$ = 1.5 * 10¹¹$*({\frac{29.5}{1})}^{2/3}$ = 1.4 X ${10}^{12}$ m