What is Escape Velocity-its derivation-examples-applications Universal Gravitational Law-Derivation and Examples Gravitational Field And Gravitational Potential-Derivation, Realation and numericals Radial Velocity and acceleration-derivation and examples Transverse Velocity and acceleration and examples

1. TOPICS:
ESCAPE
VELOCITY
GRAVITATIONAL
FIELD & POTENTIAL
UNIVERSAL
GRAVITATIONAL LAW
RADIAL & TRANSVERSAL
VELOCITY & ACCELERATION

3. ESCAPE
VELOCITY:
The minimum velocity with
which a body should be
projected upward so that it
escapes out of gravitational
force, is known as Escape
velocity from surface of Earth.

4. a spacecraft leaving the surface of Earth needs to be
going 7 miles per second, or nearly 25,000 miles per hour
to leave without falling back to the surface or falling into
orbit.
EXAMPLE:

5. DERIVATION
OF ESCAPE
VELOCITY:

6. ESCAPE VELOCITY
FORMULA:

7. “
”
CALCULATION:

8. Universal Gravitational Law:
 According to this law, Every object in the universe attracts every other
object with a force whose magnitude is proportional to the product of their
masses & inversely proportional to the square of the distance.
 Derivation
𝐅 ∝ 𝐦𝟏 𝐦𝟐 --------(1)
𝐅 ∝
𝟏
𝐫𝟐 ---------(2)
𝐅 ∝
𝐦𝟏 𝐦𝟐
𝐫𝟐 [By combining equation 1&2]
𝐅 = 𝐆
𝐦𝟏 𝐦𝟐
𝐫𝟐
 G = Universal Gravitational constant
 G = 6.673 × 10−11
N𝑚2
𝑘𝑔−2
 Its SI unit is N𝒎𝟐
𝒌𝒈−𝟐

9. • Importance of Universal Gravitational Law:
 The Gravitational force hold
the solar system together
 Holfing the atmosphere near
the surface of earth
 Motion of Planet around Sun
 Motion of Moon around
Earth
 Force that binds us to the
earth
 Tides due to moon and sun

10.  It is defined as gravitational force per unit a small test
mass.
 It is a vector Field which is denoted as “g”.
 If gravitational force F is exerted on small test mass 𝑚0
then, it is expressed as;
g =
𝐹
𝑚0
----------(a)
 Thus, gravitational field is used to measure
gravitational phenomenon and it is measured in
𝑁
𝑘𝑔
.
Gravitational Field:

11.  What is the gravitational field of object whose mass is 7. 𝟎𝒌𝒈 and its gravitational force is
1.32N ?
 Solution:
g = ?
As,
g =
𝐹
𝑚0
So,
g =
7.0
1.32
(By putting values)
g = 5.303
𝑁
𝑘𝑔
Numerical Problem:

12.  The gravitational potential at a point in a gravitational field
is defined as absolute gravitational potential energy per
unit mass at that point.
 It is a scalar point function and denoted by V(r).
Gravitational Potential

13. Derivation
 Let, U(r) be the absolute potential energy of a system having a mass M and small test
mass 𝑚0 such that M lies at origin and 𝑚0 lies at a distance r from it. Then,
U(r) = −
𝑮𝑴𝒎𝟎
𝒓
------------(1)
 If V(r) is a gravitational potential at the location 𝑚0 then,
V(r) =
𝑼(𝒓)
𝒎𝟎
------------(2)
=
𝟏
𝒎𝟎
(−
𝑮𝑴𝒎𝟎
𝒓
) [By putting values of (1) in(2)]
So,
 V(r) = −
𝑮𝑴
𝒓
------------(3)

14.  Differentiate eq (3) w.r.t. ‘r’ we get;
𝑑𝑉(𝑟)
𝑑𝑟
=
𝐺𝑀
𝑟2
 Multiply and divide by -𝑚0, so it becomes;
= −
1
𝑚0
( −
𝐺𝑀𝑚0
𝑟2 )
𝑑𝑉(𝑟)
𝑑𝑟
= −
𝐹
𝑚0
[F = ( −
𝐺𝑀𝑚0
𝑟2 )]
𝑑𝑉(𝑟)
𝑑𝑟
= -g [ g =
𝐹
𝑚0
]
∆𝑉 = −𝒈
 This show that g is negative gradient of gravitational potential.
Relation b/w Gravitational Field and
Gravitational Potential

15. Radial And Trasverse Components Of Velocity
 Point in polar coordinate system is
denoted as P(r,𝜽)
 r is the radial vector and 𝜽 is the
angle which OP line makes with
positive x-axis
 𝑟 is the unit vector along radial
vector which is called radial
direction.
 The direction perpendicular to r
called transeverse direction
denoted as 𝑠

16. Conti…
𝑣 = 𝑣r𝑟 + v𝜽𝑠
𝛚 =
𝐝𝜽
𝒅𝒕
𝒌……………eq. 1
𝑣 =
𝐝𝒓
𝒅𝜽
=
𝐝
𝒅𝜽
(𝑟. 𝑟)
=
𝐝𝒓
𝒅𝜽
𝑟 + 𝑟
𝐝𝒓
𝒅𝜽
= 𝑟𝑟 + 𝑟
𝐝𝒓
𝒅𝜽 …………eq. 2
𝑟 × 𝑠 = 𝒌
𝑠 × 𝒌= 𝑟
𝒌 × 𝑟= 𝑠
Orthogonal unit vectors

17. Conti…

𝐝𝒓
𝒅𝜽
= v = 𝛚 × 𝑟

𝐝𝒓
𝒅𝜽
= (
𝐝𝜽
𝒅𝒕
𝒌) × 𝑟

𝐝𝒓
𝒅𝜽
= 𝜽𝑠
 Put in eq. 2
 𝑣= 𝑟𝑟 + r𝜽𝑠
 Here
 𝒗r= 𝒓 & 𝐯𝜽= 𝐫𝜽

18. EXAMPLE
 Find the radial and transverse components of velocity of a particle moving
along the curve ax2 + by2 = 1 at any time t if the polar angle is 𝜽 = ct2
Solution:
Given that:
 𝜽 = ct2
Differentiate w.r.t “t”, we get
𝐝𝜽
𝒅𝒕
= 2ct
 Also given that
ax2 + by2
= 1
First we change this into polar form by putting x = rcos and y = rsin

19. Cont…
 ar2
cos2
 + br2
sin2
 = 1
 r2
(acos2
 + bsin2
) = 1
 Taking square root on both side:
 r√acos2
 + bsin2
 = 1
 r = (acos2
 + bsin2
)
−1/2
 Differentiate w.r.t “t”, we get
𝐝𝒓
𝒅𝒕
= −
1
2
(acos2  + bsin2 )−
3
2
(−a2cossin
𝐝𝜽
𝒅𝒕
+ b2sincos
𝐝𝜽
𝒅𝒕

20. Conti…

𝐝𝒓
𝒅𝒕
=
1
2
(acos2  + bsin2 )
−
3
2 (a − b)sin2
𝐝𝜽
𝒅𝒕

𝐝𝒓
𝒅𝒕
=
1
2
(acos2  + bsin2 )
−
3
2 (a − b)sin2.2ct

𝐝𝒓
𝒅𝒕
=
ct(a − b)sin2
(acos2
 + bsin2
) 3/2
 Radial component of velocity =
𝐝𝒓
𝒅𝒕
=
ct(a − b)sin2
(acos2
 + bsin2
) 3/2
Transverse component of velocity= r
𝐝𝜽
𝒅𝒕 =
(acos2
 + bsin2
)
−1/2 2ct

21. Radial and Transverse Components
of acceleration
 Let 𝑎 be the acceleration, Then
 𝑎= 𝑑𝑣
𝑑𝑡

𝑑
𝑑𝑡
(𝑟𝑟 + r𝜃𝑠)

𝑑𝑟
𝑑𝜃
𝑟 + 𝑟𝑑𝑟
𝑑𝑡
+ 𝑑𝑟
𝑑𝑡
𝜃𝑠 + r𝑑𝜃
𝑑𝑡
𝑠 + r𝜃𝑑𝑠
𝑑𝑡
→ 𝑑𝑟
𝑑𝑡
= 𝜃𝑠
 𝑟𝑟 + 𝑟𝜃𝑠 + 𝑟𝜃𝑠 + r𝜃𝑠 + r𝜃𝑑𝑠
𝑑𝑡
 As we know that,
𝑑𝑠
𝑑𝑡
= ω×𝑠 = 𝜃𝑘×𝑠 = -𝜃𝑟 𝑠 × 𝑘 = 𝑟

22. Conti.....
 𝑟𝑟 + 𝑟𝜃𝑠 + 𝑟𝜃𝑠 + r𝜃𝑠 + (- r𝜃𝜃𝑠)
 𝑟𝑟 + 2𝑟𝜃𝑠 + r𝜃𝑠 - 𝜃2r𝑟
 Then,
 𝑟(𝑟 − 𝜃2r) + 𝑠(2𝑟𝜃 + r𝜃)
 Here,
 Radial and Transverse acceleration is,
 𝑎 = 𝑎r𝑟 + 𝑎𝜃𝑠 & 𝑎𝜃 = 2𝑟𝜃 + r𝜃

23. For Example:
 Find the radial and transverse components of acceleration of
a particle moving along the circle 𝒙𝟐
+ 𝒚𝟐
= 𝒂𝟐
with constant
velocity c.
Given that
𝑑𝜃
𝑑𝑡
= c
Differentiate w.r.t “t”, we get
𝑑2
𝑑𝑡2𝜃 = 0
Also given that
𝑥2 + 𝑦2 = 𝑎2
First we change this into polar form by putting x=rcos𝜃 and y=rsin𝜃

24. Conti.....
 𝑟2𝑐𝑜𝑠2 𝜃 + 𝑟2𝑠𝑖𝑛2𝜃 = 𝑎2
 𝑟2
(𝑐𝑜𝑠2
𝜃 + 𝑠𝑖𝑛2
𝜃) =𝑎2
 𝑟2 = 𝑎2
 Taking square root on both sides
 r = a

𝑑𝑟
𝑑𝑡
= 0 → 𝑑2
𝑑𝑡2r = 0
 Radial component of acceleration
 𝑎𝑟=
𝑑2𝑟
𝑑𝑡2 - r(𝑑𝜃
𝑑𝑡
)2

25. Cont.……
 = 0 - a𝑐2
 = - a𝑐2
 Transverse component of acceleration
 𝑎𝜃 = 2 𝑑𝑟
𝑑𝑡
(𝑑𝜃
𝑑𝑡
) + r 𝑑2𝜃
𝑑𝑡2
 = 0