Gravitation

This is a  Short Answer Type Questions as classified in NCERT Exemplar

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

This is a  Short Answer Type Questions as classified in NCERT Exemplar

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.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

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 hen

This is a Long Type Questions as classified in NCERT Exemplar

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

This is a Long Type Questions as classified in NCERT Exemplar

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

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

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

This is a Long Type Questions as classified in NCERT Exemplar

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/

Initial kinetic energy of the rocket = $\frac{1}{2}{mv}^2$

Initial potential energy of the rocket = $\frac{-GMm}{R}$

Total initial energy = $\frac{1}{2}{mv}^2-\frac{GMm}{R}$

If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

Total initial energy available = 0.8