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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.
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 πΉ1 , πΉ2 , πΉ3 β¦
...
Note: Force is a vector quantity add it vectorially
...
β’
Acceleration due to gravity on the surface of planet is given by
π=
π=
π=
πΊπ
(negative rotation effect)
π 2
πΊ
4
( ππ 3 ) π
π 2 3
4πΊπππ
3
π β ππ
Points to remember: β’
Acceleration dur to gravity on the surface of earth is 9
...
8 π/π 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 π due to rotation
Solution:
Step 1: Variation of β²πβ² 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 β²π β² from center of shell is shown below
...
β’
Graph between πΈ and distance β²π β² 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
π=β
πΊπ
π
(π 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
ππ = β
2πΊπ
π
= β2ππ
8
ππ = π β ππΊπ
3
Step 1: Calculation of potential energy
β’
Potential energy = mass Γ potential
π = ππ
β’
Calculate potential using potential templates
Step 2: calculation of total energy
β’
Total energy
ππΈ = π + πΎπΈ
(πΎπΈ is kinetic energy)
1
πΎπΈ = ππ 2
2
β’
For escape velocity we can write
ππΈ = 0
Substitute the values to find the escape velocity
Question 16:
Relation between potential and field
β’
Change potential Ξπ is given by
β β
Ξπ = β β« πΈ β ππ
and
β
ππ£
πΈ =β
ππ
Step 1: calculation of gravitational field from potential
β’
If β²πβ² is the potential function as a function of π₯, π¦, π§ then field is given by
β§
β
ππ β§
ππ β§
ππ
πΈ = β ( π + π + π)
ππ₯
β’
ππ¦
ππ§
β
Substitute the value of "π" to find the value of πΈ
Question 17:
Circular Orbits
Solution:
Step 1: calculation of orbital speed
β’
Speed of a satellite in circular orbit is given by
π0 = β
πΊπ
π
π is mass of planet
π is radius of the orbit
β’
Orbital speed doesnβt depend on mass of satellite
Step 2: Calculation angular velocity and time period
β’
Angular velocity of satellite is given by
π=β
β’
πΊπ
π3
Time period is given by
π=
2π
π
π = 2πβ
π3
πΊπ
Step 3: calculation of potential energy and kinetic energy
β’
Potential energy of satellite in circular orbit is
π=β
β’
πΊππ3
(ππ is mass of satellite)
π
Kinetic energy of satellite in circular orbit is
1
πΎπΈ = ππ π12
2
πΎπΈ =
πΊπππ
2π
Step 4: calculation of total energy of satellite
β’
Total energy of satellite in circular orbit is
ππΈ = π + πΎπΈ
ππΈ = β
ππΈ = β
πΊπππ
π
+
πΊπππ
2π
πΊπππ
2π
Question 18:
Work done by external agent
Solution:
β’
Work done by external force system is
πππ₯π‘ = change in mechanical energy
πππ₯π‘ = Ξ(π + πΎ)
πππ₯π‘ = Ξπ + ΞπΎ
Step 1: Calculation of initial and final potential energies
β’
We know that potential energy π = ππ
β’
Use potential and potential energy templates to find initial and final potential energies
β’
Calculate change in potential energy Ξπ
Ξπ = ππ β ππ
Step 2: Calculation of change in kinetic energy
β’
Calculate change in kinetic energy of system
ΞπΎπΈ = πΎπ β πΎπ
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
πππ₯π‘ = Ξπ + ΞπΎ
πππ₯π‘ = (ππ + πΎπ ) β (ππ + πΎπ )
πππ₯π‘ = ππΈπ β ππΈπ
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
πΏπ = πΏπ
β’
If net force on system is zero linear momentum is conserved
β β
ππ = ππ
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 Keplerβs second law
β’
According to Keplerβs 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
πΊπ 1βπ
ππππ = β
β’
π
(
1+π
)
Velocity of planet in elliptical orbit is maximum at perigee
ππππ₯ = β
πΊπ 1βπ
π
(
1+π
)
ππππ (1 + π) = ππππ₯ (1 β π)
Step 2: Calculation of total energy in elliptical orbit
β’
Total energy of a satellite revolving in elliptical orbit is constant
ππΈ = β
πΊπππ
2π
(π is semi major axis)
ππ 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.
Description: Provide gravitation questions and answers with easy explanations.