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Question 1:
Universal law of gravitation

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

→ → →
If 𝐹1 , 𝐹2 , 𝐹3 …
...


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

•

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

𝐺𝑀

(negative rotation effect)

𝑅2
𝐺

4

( 𝜋𝑅3 ) 𝜌

𝑅2 3

4𝐺𝜌𝜋𝑅
3

𝑔 ∝ 𝜌𝑅
Points to remember: •

Acceleration dur to gravity on the surface of earth is 9
...
8 𝑚/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 ′𝒈′ due to rotation of earth
...


Step 2: Calculation of net field due to multiple point masses
•
•
•

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 ′𝑟 ′ from center of shell is shown below
...


•

Graph between 𝐸 and distance ′𝑟 ′ 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
•

Potential due to point mass is given by
𝑉=−

𝐺𝑀
𝑟

(𝑟 is distance from mass)

Step 2: Calculation of potential due to system of point masses
•

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

or

•

The velocity for which total energy of object is zero

•

Escape velocity form the surface of planet is
𝑉𝑒 = √

2𝐺𝑀
𝑅

= √2𝑔𝑅

8

𝑉𝑒 = 𝑅√ 𝜋𝐺𝜌
3

Step 1: Calculation of potential energy
•

Potential energy = mass × potential
𝑈 = 𝑚𝑉

•

Calculate potential using potential templates

Step 2: calculation of total energy

•

Total energy
𝑇𝐸 = 𝑈 + 𝐾𝐸

(𝐾𝐸 is kinetic energy)

1

𝐾𝐸 = 𝑚𝑉 2
2

•

For escape velocity we can write
𝑇𝐸 = 0

Substitute the values to find the escape velocity

Question 16:
Relation between potential and field
•

Change potential Δ𝑉 is given by
→ →
Δ𝑉 = − ∫ 𝐸 ⋅ 𝑑𝑟
and
→
𝑑𝑣
𝐸 =−
𝑑𝑟

Step 1: calculation of gravitational field from potential
•

If ′𝑉′ is the potential function as a function of 𝑥, 𝑦, 𝑧 then field is given by
∧
→
𝜕𝑉 ∧
𝜕𝑉 ∧
𝜕𝑉
𝐸 = − ( 𝑖 + 𝑗 + 𝑘)
𝜕𝑥

•

𝜕𝑦

𝜕𝑧

→
Substitute the value of "𝑉" to find the value of 𝐸

Question 17:
Circular Orbits

Solution:
Step 1: calculation of orbital speed
•

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

𝐺𝑀
𝑟

𝑀 is mass of planet
𝑟 is radius of the orbit
•

Orbital speed doesn’t depend on mass of satellite

Step 2: Calculation angular velocity and time period
•

Angular velocity of satellite is given by
𝜔=√

•

𝐺𝑀
𝑟3

Time period is given by
𝑇=

2𝜋
𝜔

𝑇 = 2𝜋√

𝑟3
𝐺𝑀

Step 3: calculation of potential energy and kinetic energy
•

Potential energy of satellite in circular orbit is
𝑈=−

•

𝐺𝑀𝑚3

(𝑚𝑠 is mass of satellite)

𝑟

Kinetic energy of satellite in circular orbit is
1

𝐾𝐸 = 𝑚𝑠 𝑉12
2

𝐾𝐸 =

𝐺𝑀𝑚𝑠
2𝑟

Step 4: calculation of total energy of satellite
•

Total energy of satellite in circular orbit is
𝑇𝐸 = 𝑈 + 𝐾𝐸
𝑇𝐸 = −
𝑇𝐸 = −

𝐺𝑀𝑚𝑠
𝑟

+

𝐺𝑀𝑚𝑠
2𝑟

𝐺𝑀𝑚𝑠
2𝑟

Question 18:
Work done by external agent

Solution:
•

Work done by external force system is
𝜔𝑒𝑥𝑡 = change in mechanical energy
𝜔𝑒𝑥𝑡 = Δ(𝑈 + 𝐾)

𝜔𝑒𝑥𝑡 = Δ𝑈 + Δ𝐾
Step 1: Calculation of initial and final potential energies
•

We know that potential energy 𝑈 = 𝑚𝑉

•

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

•

Calculate change in potential energy Δ𝑈
Δ𝑈 = 𝑈𝑓 − 𝑈𝑖

Step 2: Calculation of change in kinetic energy
•

Calculate change in kinetic energy of system
Δ𝐾𝐸 = 𝐾𝑓 − 𝐾𝑖

•

If velocity is zero kinetic energy is zero

Step 3: Calculation of work done by external agent
•

Work done by external agent is given by
𝜔𝑒𝑥𝑡 = Δ𝑈 + Δ𝐾
𝜔𝑒𝑥𝑡 = (𝑈𝑓 + 𝐾𝑓 ) − (𝑈𝑖 + 𝐾𝑖 )
𝜔𝑒𝑥𝑡 = 𝑇𝐸𝑓 − 𝑇𝐸𝑖

Question 19:
Conservation of Energy , momentum and angular momentum
•

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

Solution:
Step 1: Calculation of potential energy and kinetic energy
•

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


•

At maximum height kinetic energy is zero
...


•

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

•

If net force on system is zero linear momentum is conserved
→ →
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
•

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


•

Velocity of planet in elliptical orbit is minimum at apogee
𝐺𝑀 1−𝑒

𝑉𝑚𝑖𝑛 = √
•

𝑎

(

1+𝑒

)

Velocity of planet in elliptical orbit is maximum at perigee
𝑉𝑚𝑎𝑥 = √

𝐺𝑀 1−𝑒
𝑎

(

1+𝑒

)

𝑉𝑚𝑖𝑛 (1 + 𝑒) = 𝑉𝑚𝑎𝑥 (1 − 𝑒)
Step 2: Calculation of total energy in elliptical orbit
•

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

𝐺𝑀𝑚𝑝
2𝑎

(𝑎 is semi major axis)

𝑚𝑝 is mass of planet
Points to remember: •

In elliptical orbit mechanical energy and angular momentum is conserved

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