Gravitational field and potential, escape velocity, universal gravitational law, radia and transverse velocity and acceleration with derivation and examples
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
Similar to Gravitational field and potential, escape velocity, universal gravitational law, radia and transverse velocity and acceleration with derivation and examples
PHYSICS - Rotational dynamics (MAHARASHTRA STATE BOARD)Pooja M
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Similar to Gravitational field and potential, escape velocity, universal gravitational law, radia and transverse velocity and acceleration with derivation and examples (20)
Gravitational field and potential, escape velocity, universal gravitational law, radia and transverse velocity and acceleration with derivation and examples
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:
8. Universal Gravitational Law:
๏ต 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.
๏ต Derivation
๐ โ ๐ฆ๐ ๐ฆ๐ --------(1)
๐ โ
๐
๐ซ๐ ---------(2)
๐ โ
๐ฆ๐ ๐ฆ๐
๐ซ๐ [By combining equation 1&2]
๐ = ๐
๐ฆ๐ ๐ฆ๐
๐ซ๐
๏ต G = Universal Gravitational constant
๏ต G = 6.673 ร 10โ11
N๐2
๐๐โ2
๏ต Its SI unit is N๐๐
๐๐โ๐
9. โข Importance of Universal Gravitational Law:
๏ต The Gravitational force hold
the solar system together
๏ต Holfing the atmosphere near
the surface of earth
๏ต Motion of Planet around Sun
๏ต Motion of Moon around
Earth
๏ต Force that binds us to the
earth
๏ต 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 โgโ.
๏ If gravitational force F is exerted on small test mass ๐0
then, it is expressed as;
g =
๐น
๐0
----------(a)
๏ Thus, gravitational field is used to measure
gravitational phenomenon and it is measured in
๐
๐๐
.
Gravitational Field:
11. ๏ถ What is the gravitational field of object whose mass is 7. ๐๐๐ and its gravitational force is
1.32N ?
๏ถ Solution:
g = ?
As,
g =
๐น
๐0
So,
g =
7.0
1.32
(By putting values)
g = 5.303
๐
๐๐
Numerical Problem:
12. ๏ฑ The gravitational potential at a point in a gravitational field
is defined as absolute gravitational potential energy per
unit mass at that point.
๏ฑ It is a scalar point function and denoted by V(r).
Gravitational Potential
13. Derivation
๏ฑ Let, U(r) be the absolute potential energy of a system having a mass M and small test
mass ๐0 such that M lies at origin and ๐0 lies at a distance r from it. Then,
U(r) = โ
๐ฎ๐ด๐๐
๐
------------(1)
๏ฑ If V(r) is a gravitational potential at the location ๐0 then,
V(r) =
๐ผ(๐)
๐๐
------------(2)
=
๐
๐๐
(โ
๐ฎ๐ด๐๐
๐
) [By putting values of (1) in(2)]
So,
๏ฑ V(r) = โ
๐ฎ๐ด
๐
------------(3)
14. ๏ถ Differentiate eq (3) w.r.t. โrโ we get;
๐๐(๐)
๐๐
=
๐บ๐
๐2
๏ถ Multiply and divide by -๐0, so it becomes;
= โ
1
๐0
( โ
๐บ๐๐0
๐2 )
๐๐(๐)
๐๐
= โ
๐น
๐0
[F = ( โ
๐บ๐๐0
๐2 )]
๐๐(๐)
๐๐
= -g [ g =
๐น
๐0
]
โ๐ = โ๐
๏ 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
๏ต Point in polar coordinate system is
denoted as P(r,๐ฝ)
๏ต r is the radial vector and ๐ฝ is the
angle which OP line makes with
positive x-axis
๏ต ๐ is the unit vector along radial
vector which is called radial
direction.
๏ต The direction perpendicular to r
called transeverse direction
denoted as ๐
18. EXAMPLE
๏ต 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 ๐ฝ = ct2
Solution:
Given that:
๏ต ๐ฝ = ct2
Differentiate w.r.t โtโ, we get
๐๐ฝ
๐ ๐
= 2ct
๏ต Also given that
ax2 + by2
= 1
First we change this into polar form by putting x = rcos๏ฑ and y = rsin๏ฑ
23. For Example:
๏ฑ Find the radial and transverse components of acceleration of
a particle moving along the circle ๐๐
+ ๐๐
= ๐๐
with constant
velocity c.
Given that
๐๐
๐๐ก
= c
Differentiate w.r.t โtโ, we get
๐2
๐๐ก2๐ = 0
Also given that
๐ฅ2 + ๐ฆ2 = ๐2
First we change this into polar form by putting x=rcos๐ and y=rsin๐