Physics NCERT Exemplar Solutions Class 11th Chapter Eight: Overview, Questions, Preparation

Explanation- when a body og mass m is revolving around a star of mass M.

Linear velocity of the body  v= $\sqrt[]{\frac{GM}{r}}$  so when r increases then v decreases.

Angular velocity of the body w = 2 $\pi /T$

According to kepler’s law T² $\propto$ r³

 So T= kr^3/2

So w= $\frac{2\pi }{k{r}^{3/2}}$ so when r increases , w decreases.

Kinetic energy of the body K= 1/2mv²=1/2m( $\frac{GM}{r}$ )  so when we increase r ,KE decreases.

Gravitational potential energy of the body

U=-GMm/r

So when we increase r, PE becomes less negative

Total energy of the body E= KE+PE= $\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}$

When r increases total energy becomes less negative . i.e increases.

Angular momentum of the body L =mvr =mr $\sqrt[]{\frac{GM}{r}}=$ m $\sqrt[]{GMr}$

L= $\sqrt[]{r}$

Explanation- consider a diagram having vertices A,B,C,D,E and F

AC= AG+GC=2AG

= 2lcos30⁰= 2l $\frac{\sqrt[]{3}}{2}$

= $\sqrt[]{3}l=AE$

AD=AH+HJ+JD= lsin30⁰+l+lsin30⁰=2l

Force on mass m at A due to mass m at B is f₁= $\frac{Gmm}{l^2}$  along AB

Force on mass m at A due to mass m at C is f₂= $\frac{Gmm}{\sqrt[]{3}{l}^2}=\frac{G{m}^2}{3l^2}$  along AC

Force on mass m at A due to mass m at D is F₃= $\frac{Gmm}{({2l)}^2}=\frac{G{m}^2}{4l^2}$ along AD

Force on mass m at A due to mass m at E is F₄= $\frac{Gmm}{\sqrt[]{3}{l}^2}$ = $\frac{G{m}^2}{3l^2}$ along AE

Force on mass m at A due to mass m at F is F₅= $\frac{G{m}^2}{l^2}$  along AF

Resultant force due to F₁ and F₅ is F₁= $\sqrt[]{f_1^2+f_5^2+2f_1{f}_{2}cos60}$

= $\frac{\sqrt[]{3}G{m}^2}{3l^2}=\frac{G{m}^2}{\sqrt[]{3}{l}^2}$ along AD

So net force along AD = F₁+F₂+F₃= $\frac{G{m}^2}{l^2}+\frac{G{m}^2}{\sqrt[]{3}{l}^2}+\frac{G{m}^2}{4l^2}=\frac{G{m}^2}{l^2}\left(1+\frac{1}{\sqrt[]{3}}+\frac{1}{4}\right)$

8.3 A satellite is to be placed in equatorial geostationary orbit around earth for communication.

(a) Calculate height of such a satellite.

(b) Find out the minimum number of satellites that are needed to cover entire earth so that at least one satellites is visible from any point on the equator.

 [M = 6 × 1024 kg, R = 6400 km, T = 24h, G = 6.67 × 10^-11 SI units

Explanation- mass of earth M = 6 $\times {10}^{24}kg$

Radius of the earth , R = 6400km= 6.4 $\times {10}^{6}m$

T=24h= 24 $\times 60\times 60=86400s$

G = 6.67 $\times$ 10^-11Nm²/kg²

(a) Time period T = $2\pi \sqrt[]{\frac{{\left(R+h\right)}^3}{GM}}$

$T^2$ = $4\pi$ ^{2 $\frac{{(R+h)}^3}{GM}$}

$R+h$ = ( $\frac{T^{2}GM}{4{\pi }^2}$ )^1/3

$h=$  ( $\frac{T^{2}GM}{4{\pi }^2}$ )^1/3-R

So after solving we get h = 3.59 $\times {10}^{7}m$

(b) If satellite is at height h from the earth’s surface

Cos $\theta$ = $\frac{R}{R+h}=\frac{1}{\left(1+\frac{h}{R}\right)}=\frac{1}{1+5.61}=\frac{1}{6.61}=0.1513$ = cos81⁰18’

$\theta$ = 81⁰18’

$2\theta$ =2(81⁰18’)= 162⁰36’

If n number of satellite needed to cover entire the earth then

So n = 360⁰/2 $\theta$ = 2.31

So minimum 3 satellite are required to cover entire earth.

Explanation-let m be the mass of the earth v_p,v_a be the velocity of the earth at perigee and apogee respectively. Similarly w_p and w_a are angular velocities.

At perigee , $r_p^2{w}_p=r_a^2{w}_a$ at apogee

If a is the semimajor axis of the earth’s orbit then r_p=a(1-e) and r_a=a(1+e)

$\frac{w_p}{w_a}={\left(\frac{1+e}{1-e}\right)}^{2}e$ = 0.00167

$\frac{w_p}{w_a}=1.0691$

Let w be the angular speed which is geometric mean of w_p and w_a and corresponding to mean solar day.

$\left(\frac{w_p}{w}\right)\left(\frac{w}{w_a}\right)$ =1.0691

If w corresponds to 1⁰ per day then w_p = 1.034⁰ per day and w_a= 0.967⁰ per day . since 361⁰=24, mean solar day we get 361.034⁰ which corresponds to 24h,8.14’’ and 360.967⁰ corresponds to 23h59min52’’. So this does not explain the actual variation of the length of the day during the year.