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Title: Physics Gravitation Class 11 Questions and Their Explanations With diagrams and figures
Description: Provide gravitation questions and answers with easy explanations.

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Question 1:
Universal law of gravitation

Solution:
Step 1: Calculation of force between two masses
...

β†’ β†’ β†’
If 𝐹1 , 𝐹2 , 𝐹3 …
...


Note: Force is a vector quantity add it vectorially
...

β€’

Acceleration due to gravity on the surface of planet is given by
𝑔=
𝑔=
𝑔=

𝐺𝑀

(negative rotation effect)

𝑅2
𝐺

4

( πœ‹π‘…3 ) 𝜌

𝑅2 3

4πΊπœŒπœ‹π‘…
3

𝑔 ∝ πœŒπ‘…
Points to remember: β€’

Acceleration dur to gravity on the surface of earth is 9
...
8 π‘š/𝑠 2

Question 3:
Variation of g above the surface of the earth

Solution:
Step 1: Acceleration due to gravity above the surface of earth
...


Question 5:

Variation of 𝑔 due to rotation

Solution:
Step 1: Variation of β€²π’ˆβ€² due to rotation of earth
...


Step 2: Calculation of net field due to multiple point masses
β€’
β€’
β€’

Calculate field due to each point mass at a given point
From superposition principal, to calculate net field we add them vectorially
...

Note: (1) Field is a vector it should be added vectorially
(2) At null point field is zero
...

Variation of field as a function of distance β€²π‘Ÿ β€² from center of shell is shown below
...


β€’

Graph between 𝐸 and distance β€²π‘Ÿ β€² from center of sphere is

Question 10:
Potential due to point mass and system of point masses

Solution:
Step 1: calculation of potential due to point mass
β€’

Potential due to point mass is given by
𝑉=βˆ’

𝐺𝑀
π‘Ÿ

(π‘Ÿ is distance from mass)

Step 2: Calculation of potential due to system of point masses
β€’

If more than one mass is present, the potential at a point is the sum of potential due to individual mass
...

or

β€’

The velocity for which total energy of object is zero

β€’

Escape velocity form the surface of planet is
𝑉𝑒 = √

2𝐺𝑀
𝑅

= √2𝑔𝑅

8

𝑉𝑒 = π‘…βˆš πœ‹πΊπœŒ
3

Step 1: Calculation of potential energy
β€’

Potential energy = mass Γ— potential
π‘ˆ = π‘šπ‘‰

β€’

Calculate potential using potential templates

Step 2: calculation of total energy

β€’

Total energy
𝑇𝐸 = π‘ˆ + 𝐾𝐸

(𝐾𝐸 is kinetic energy)

1

𝐾𝐸 = π‘šπ‘‰ 2
2

β€’

For escape velocity we can write
𝑇𝐸 = 0

Substitute the values to find the escape velocity

Question 16:
Relation between potential and field
β€’

Change potential Δ𝑉 is given by
β†’ β†’
Δ𝑉 = βˆ’ ∫ 𝐸 β‹… π‘‘π‘Ÿ
and
β†’
𝑑𝑣
𝐸 =βˆ’
π‘‘π‘Ÿ

Step 1: calculation of gravitational field from potential
β€’

If ′𝑉′ is the potential function as a function of π‘₯, 𝑦, 𝑧 then field is given by
∧
β†’
πœ•π‘‰ ∧
πœ•π‘‰ ∧
πœ•π‘‰
𝐸 = βˆ’ ( 𝑖 + 𝑗 + π‘˜)
πœ•π‘₯

β€’

πœ•π‘¦

πœ•π‘§

β†’
Substitute the value of "𝑉" to find the value of 𝐸

Question 17:
Circular Orbits

Solution:
Step 1: calculation of orbital speed
β€’

Speed of a satellite in circular orbit is given by
𝑉0 = √

𝐺𝑀
π‘Ÿ

𝑀 is mass of planet
π‘Ÿ is radius of the orbit
β€’

Orbital speed doesn’t depend on mass of satellite

Step 2: Calculation angular velocity and time period
β€’

Angular velocity of satellite is given by
πœ”=√

β€’

𝐺𝑀
π‘Ÿ3

Time period is given by
𝑇=

2πœ‹
πœ”

𝑇 = 2πœ‹βˆš

π‘Ÿ3
𝐺𝑀

Step 3: calculation of potential energy and kinetic energy
β€’

Potential energy of satellite in circular orbit is
π‘ˆ=βˆ’

β€’

πΊπ‘€π‘š3

(π‘šπ‘  is mass of satellite)

π‘Ÿ

Kinetic energy of satellite in circular orbit is
1

𝐾𝐸 = π‘šπ‘  𝑉12
2

𝐾𝐸 =

πΊπ‘€π‘šπ‘ 
2π‘Ÿ

Step 4: calculation of total energy of satellite
β€’

Total energy of satellite in circular orbit is
𝑇𝐸 = π‘ˆ + 𝐾𝐸
𝑇𝐸 = βˆ’
𝑇𝐸 = βˆ’

πΊπ‘€π‘šπ‘ 
π‘Ÿ

+

πΊπ‘€π‘šπ‘ 
2π‘Ÿ

πΊπ‘€π‘šπ‘ 
2π‘Ÿ

Question 18:
Work done by external agent

Solution:
β€’

Work done by external force system is
πœ”π‘’π‘₯𝑑 = change in mechanical energy
πœ”π‘’π‘₯𝑑 = Ξ”(π‘ˆ + 𝐾)

πœ”π‘’π‘₯𝑑 = Ξ”π‘ˆ + Δ𝐾
Step 1: Calculation of initial and final potential energies
β€’

We know that potential energy π‘ˆ = π‘šπ‘‰

β€’

Use potential and potential energy templates to find initial and final potential energies

β€’

Calculate change in potential energy Ξ”π‘ˆ
Ξ”π‘ˆ = π‘ˆπ‘“ βˆ’ π‘ˆπ‘–

Step 2: Calculation of change in kinetic energy
β€’

Calculate change in kinetic energy of system
Δ𝐾𝐸 = 𝐾𝑓 βˆ’ 𝐾𝑖

β€’

If velocity is zero kinetic energy is zero

Step 3: Calculation of work done by external agent
β€’

Work done by external agent is given by
πœ”π‘’π‘₯𝑑 = Ξ”π‘ˆ + Δ𝐾
πœ”π‘’π‘₯𝑑 = (π‘ˆπ‘“ + 𝐾𝑓 ) βˆ’ (π‘ˆπ‘– + 𝐾𝑖 )
πœ”π‘’π‘₯𝑑 = 𝑇𝐸𝑓 βˆ’ 𝑇𝐸𝑖

Question 19:
Conservation of Energy , momentum and angular momentum
β€’

If only conservation forces are doing work, mechanical energy is conserved

Solution:
Step 1: Calculation of potential energy and kinetic energy
β€’

Calculate initial and final potential energy using potential and potential energy templates
...


β€’

At maximum height kinetic energy is zero
...


β€’

If net torque about a point is zero angular momentum can be conserved about that point (generally center of planet)
β†’
β†’
𝐿𝑖 = 𝐿𝑓

β€’

If net force on system is zero linear momentum is conserved
β†’ β†’
𝑃𝑖 = 𝑃𝑓

Note: (1) At closest distance velocity along line joining the masses is zero, but object can have perpendicular
component of velocity

Question 20:
Kepler laws

Solution:
Step 1: Application of Kepler’s second law
β€’

According to Kepler’s second law, the line that joins sun and planet sweeps equal areas in equal interval of time
...


β€’

Velocity of planet in elliptical orbit is minimum at apogee
𝐺𝑀 1βˆ’π‘’

π‘‰π‘šπ‘–π‘› = √
β€’

π‘Ž

(

1+𝑒

)

Velocity of planet in elliptical orbit is maximum at perigee
π‘‰π‘šπ‘Žπ‘₯ = √

𝐺𝑀 1βˆ’π‘’
π‘Ž

(

1+𝑒

)

π‘‰π‘šπ‘–π‘› (1 + 𝑒) = π‘‰π‘šπ‘Žπ‘₯ (1 βˆ’ 𝑒)
Step 2: Calculation of total energy in elliptical orbit
β€’

Total energy of a satellite revolving in elliptical orbit is constant
𝑇𝐸 = βˆ’

πΊπ‘€π‘šπ‘
2π‘Ž

(π‘Ž is semi major axis)

π‘šπ‘ is mass of planet
Points to remember: β€’

In elliptical orbit mechanical energy and angular momentum is conserved
Title: Physics Gravitation Class 11 Questions and Their Explanations With diagrams and figures
Description: Provide gravitation questions and answers with easy explanations.