Physics NCERT Exemplar Solutions Class 11th Chapter Eight

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

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

T=24h= 24 $*60*60=86400s$

G = 6.67 $*$ 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 $*{10}^{7}m$

(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.

This is a Long Type Questions as classified in NCERT Exemplar

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= $\frac{G{m}^2}{l^2}+\frac{G{m}^2}{\sqrt[]{3}{l}^2}+\frac{G{m}^2}{4l^2}=\frac{G{m}^2}{l^2}\left(1+\frac{1}{\sqrt[]{3}}+\frac{1}{4}\right)$