Physics NCERT Exemplar Solutions Class 11th Chapter Eight

Explanation- A body can be shielded from the gravitational influence of nearby matter because gravitational force between two point mass bodies is independent of the intervening medium. we cannot shield a body from gravitational influence of nearby matter.

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Explanation- Yes, a body can have inertia but no weight. When we taken body to the centre of earth its has some inertia but no weight .

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Explanation- Gravitational forces acting between two point masses m₁ and m₂, F= Gm₁m₂/r² is independent of nature between them. So the gravitational force will remain unaffected when dipped in water.

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Explanation-areal velocity of earth and sun

dA/dt=L/2m

where L is angular momentum

but we know L= r $*P=r\left(mv\right)$

dA/dt=1/2m (r $*mv$ )=1/2 (r $*v$ )

so dA/dt direction will be perpendicular to the plane containing r and v

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Explanation-

areal velocity of a planet revolving around the sun is constant with respect to time.

Explanation- examples of central force – gravitational force and electrostatic force

Examples of non central force = nuclear force, magnetic force acting between two current carrying loops etc.

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Explanation-air molecules in the atmosphere are attracted vertically downward by the gravitational force of the earth just like an apple falling from the tree. Air molecules move randomly due to their thermal velocity and hence resultant motion of air molecules move randomly in vertical downward direction. But in case of apple it is heavier than air so it fall downwards only.

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Explanation- r_p= radius of perihelion =2R

R_a = radius of aphelion =6R

So r_a=a(1+e)=6R

And r_p= a(1-e)=2R

Solving above eqns

 We get e = ½

By conservation of angular momentum angular momentum at perigee= angular momentum at apogee

So mv_pr_p=mv_ar_a

V_a/v_p=1/3

Where m is mass of satellite .

By conservation of energy , energy at perigee =energy at apogee

$\frac{1}{2}m{v}_p^2$ - $\frac{GMm}{r_p}=\frac{1}{2}m{v}_a^2-\frac{GMm}{r_a}$

So $v_p^2\left(1-\frac{1}{9}\right)=-2GM\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=2GM\left(\frac{1}{r_p}-\frac{1}{r_a}\right)$

V_p= $\frac{{\left[\frac{2GM}{R}\right(\frac{1}{2}-\frac{1}{6}\left)\right]}^{1/2}}{[{{[1-\frac{v_a}{v_p}]}^2]}^{1/2}}$ = $\sqrt[]{\frac{3}{4}\frac{GM}{R}}=6.85km/s$

V_p=6.85km/s , v_a=2.28km/s

So orbital velocity v_c= $\sqrt[]{\frac{GM}{r}}$

R=6R ,v_c= 3.23km/s

Hence to transfer to a circular orbit at apogee , we have to boost the velocity by 3.23-2.28=0.95km/s.

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.