Class 11-science H C VERMA Solutions Physics Chapter 11 - Gravitation

F= 

F = Gravitational force between two bodies

G = Gravitational constant

m_1, m₂ = Masses of the two bodies

r = Distance between the bodies

Here, m₁=m₂=10 kg

G=6.67×10^-11 

F =  

F = 6.67×10^-7N

=  (2)=2k

Here and are on the same straight line.

Thus, Resultant force= -[ because >]

 = 6K - 2k = 4K

 and are on the same straight line.

Resultant force= -  [because >]

= 8K - 4k = 4K

Total resultant force =  

Here cos𝜃 = 0 because 𝜃 = 90° 

 =  

Resultant force on 'm' is= 4 

Mass 'm' is at mid-point of side PR.

(a) OP = OR = a/2

=  K=  

=  = 4K

= = 4K

Altitude OQ =  

=  =  =  K

Here, F_OP and F_OR are equal in magnitude and opposite in direction.

So, F_OP + F_OR = 0

Total resultant force(F)

= + + 

= 0 +  

=  K

=  (Along OQ)

(b) If particle at centroid, All forces 'F' are equal in magnitude.

F= F_GP = F_GQ = F_GR =  

Resultant of F_GQ and F_GR

F_{R =}  

= 

F_R =F

The resultant is 0, because they have equal magnitude and are in opposite direction.

Let us assume that the center of mass of spheres are located at point P, Q and R. So, re-drawing the figure, it is an equilateral triangle.

Force on mass M at point P

F_PR = F_PQ =  = k (let)

Resultant

F_{R =}  

F_R = 

F_R= 

F_R = 

Here,  

Now, force on particle 1

Due to particle 4:  

Due to particle 2:  

Due to particle 1:  

Resultant of  

 (along  direction)

Net resultant force,  

Since particle is revolving in a circle. So, this is centripetal force.

The acceleration due to gravity at a point at height 'h' from the surface of moon of radius R is

g = 

=  

= 0.65 m/s²

Initially,

Finally,

If both bodies are taken as system.

Now, applying law of conservation of energy,

On solving,

m/s

and

m/s

Consider two small elements at an angle 𝜃 from horizontal length of the element =Rd𝜃 

Mass of each element dm = (Rd𝜃)

Force by elemental mass on particle

dF =  

dF =  

dF =  

Resultant of the both elemental forces = 2dFsinθ

Net Force

F_N = 

= sinθ

= 

=  

F_{N =}  ( because  

Consider two small elements of length dx at a distance x from center on both sides of center.

Mass of each element, dm = dx

Field due to each element at point P

= 

dE =  

Resultant of the elemental field= 2dE cosθ

Net electric force =  

= 

=  

On integrating,

E_N =  

Here,  

The gravitational force of mass m due to shell of  is zero as gravitational field inside shell is zero.

Gravitational field due to shell of mass  outside the shell at mass m= 

So, force  

Method 1:

The gravitational field inside a sphere of radius R at a distance x from center is  

So, force on particle of mass m is

Method 2:

Acceleration due to gravity at depth d from surface

Here,  

So, force on particle of mass m is= mg

Distance of the particle from earth's center

Force on the particle=massgravitational field intensity

Force exerted by wall on particle=Force exerted by particle on wall

(a)

Gravitational field intensity inside shell is zero. So, force by shell on mass  is zero.

Now, for solid sphere.

Gravitational field inside sphere at a distance (x-r) from center is

So, force by sphere on mass  is

Net force

(b)

Gravitational field intensity inside shell is zero. So, force by shell on mass  is zero.

Now, for solid sphere.

Gravitational field intensity outside sphere at distance (x-r) is given by

So, force by sphere on mass  is

Net force

(c)

Force=massGravitational field intensity

Force by shell= (towards center)

Force by sphere= (towards center)

(a) At point  

Net gravitational field=field due to shell+ field due to sphere

(b) At point  

Net gravitational field=field due to shell+ field due to sphere

Resultant field inside any spherical shell is zero at all points. So, at point a and B, field is zero. The field due to upper part and lower part and are equal and opposite.

Let mass 0.5kg is placed at distance x from 2kg between both masses.

Since, particle is in equilibrium.

Force due to 2kg= Force due to 4kg

On solving,

x=0.83m

Now, gravitational potential energy

J

Work done

=ΔU = final P.E - initial P.E

=U_f -U_i

=3() - 3()

=()

Work done against gravitational force = change in potential energy

J

(a) Force,  

N

(b)  

at (0,0), potential =0

C=0

So, V=-5x-12y

At point (12,0)

At point (0,5)

(c) Potential at (12,5)

and potential at (0, 0) is 0.

So, change in gravitational potential energy

(d)  

Dimension of L.H.S =[V]

Dimension of RHS= 

Hence, dimensionally correct.

(b)  

(c)  

Gravitational field in the region is given by

E = 2 î + 3 

Slope of gravitational field, m₁ = tanθ₁ = 3/2

The line 3 y + 2 x = 5 can be written with the slope, m₂ = ₂ = -2/3

m₁ × m₂ = -1 

Thus, m₁, m₂ lines are perpendicular.

So, no work is done in by gravitational force when a particle is moved on the given line. 

Let height be h

 =  

2=  

h = (-1)R

g'=g (1- )

g' be the acceleration due to gravity at Mount Everest

h=8848

g'=9.8×(1- )

= 9.77 m/s² 

g'= g(1- )

g' be the acceleration due to gravity at surface

= 9.8×(1- )

=9.79 m/s² 

Let g_p be the gravity at the poles ,g_e be the gravity at the equator

g_e= g_p - ω²R 

=9.81 -  

= 9.766 N/m²

Now, mg_e = 1kg × 9.766 N/m²

=9.766 N

Thus the body weighs 9.766 N at the equator.

At equator, let stretch in spring be x

Spring force= 

 -(1)

Now, at height h above surface, stretch in spring is x

Spring force= 

 -(2)

From (1) and (2)

Km

Apparent acceleration due to gravity at equator becomes 0.

G'=g - ω²R = 0

g = ω²R 

ω=  

= 

= 1.2rad/s

Time period = T =  =  = 1.41 hrs 

(a) Since, ship is at rest with respect to water.

So, Angular speed of ship=Angular speed of earth

(b) The tension in string is given by

(c) If ship sails at speed v, then tension

  [Here R>>>V ]

Kepler's law of planetary motion says ,

T² ∝ R³

 =  

 =  = 1.52 

Time period of the moon around the earth is given by,

T =2π 

m = mass of earth

r = Distance between center of the moon and earth

27.3 × 86400= 2×3.14 

M = kg 

Time period of revolution of satellite around the mars is given by

T =2π 

M = Mars mass

r = Distance of the satellite from center of the planet

27540 = 2×3.14 

M = 6.5 × 10²³kgs 

(a) Speed of satellite in its orbit

V₀= 

V₀= = 6.9 km/sec

(b) Kinetic energy of satellite = (½)×1000× 

= 2.38× 

(c) Potential energy of satellite = P.E =  

= 

= -4.76 J

(d) Time period of the satellite

T=7645.2 sec

 =2.1 hrs 

(a) Since, angular speed of satellite is equal to angular speed of earth.

So, time period of satellite=24hrs

T=24hrs

h=42300 Km

(b) Time taken from North pole to equatorial plane

hrs

Weight at North pole = 10 N

W = mg

10 = mg

Weight of body in a satellite at height 'h' is

W_S =mg'

= m [g (]

For a geo-stationary satellite, R = 6.4×10³ km

h = 3.6 ×10³ km

N 

Time period of the revolution of the satellite around a planet in terms of the radius of the orbit of the satellite is given by:

T =2π 

T² =4 π² ( 

g = 

Where 'g' is acceleration due to gravity. 

Consider that 'B' is the position of the geo- stationary satellite.

Here R= 6400 kms ( Radius of earth)

d = altitude

θ = colatitude of the plane which can directly receive a signal from the satellite.

Take ∆OAB,

cos 𝜙 =  =  

𝜙 =  

= (0.15)

Now θ + 𝜙 =  

θ =  -  

θ = 

Upon Earth's surface, its total energy is

E = K.E +P.E

= mv² +(- (1)

In space at maximum height , its K.E =0 and P.E =- 

E_s = 0+  (2)

Total energy must be conserved , therefore (1) = (2)

 mv² +(- =  

Here h= R, so

 mv² =  (  

 mv² =  

V=  

= 

= 7.9 km/s

Velocity of particle on the Earth surface = V_e = 15 km/s

Let V_s be the particle velocity in space.

Total Energy must be conserved.

Change in K.E. = Change in P.E.

∆K.E.=∆P.E.

 m() = 

[(225) - ] =  

On solving

V_s  = 10 km/s 

Escape velocity = V_e = 

Mass = 6× 10²⁴ kgs.

V_e = 3×10⁸ m/s.

i.e., 3×10⁸=  

On solving,

R = 8.89×10^-3

=9×10^-3

= 9mm.