Relation between Newton’s Law of gravitation and Kepler’s Law
&
➔
=Constant
E.g. Given that the period of revolution of the earth round the sun is approximately . The distance from the earth to the sun is , estimating the mass of the sun.
Solution
Also
Since
Gravitational field
It is defined as a region any mass enters the field will experience a
Gravitational force.
Strength of field
The strength of a gravitational field is defined as force acting on unit
mass placed in the field
Field strength =
At the earth surface: field strength =
We can interpreter the meaning of g
(1) Acceleration due to gravity ()
(2) The earth gravitational field strength ()
The relation of g with law of gravitational
The force of gravitational is
➔ ➔
Gravitational mass and inertial mass
- For F = ma, m is called inertial mass. It measures the opposition of
Body to change of motion i.e. inertial.
- For if m is defined in this way , it is called gravitational
Mass.
Variation of g with latitude
The true weight = gravitational attraction
I.e. W = mg
- At poles
T - mg = 0
m= mg
g =
-
At the equator
mg - T = m
T = mg - m
m= mg - m
= g - (< g )
-
At latitude θ
The body describes a circle of radius r centre
Note:
-
Note that mCosθ is the resultant force of mg and m.
The direction of (or T) is not exactly towards the centre of the earth
except at the poles and the equator
2) Practical Values
=====================================
Latitude / m
9.78
9.79
9.82
9.83
=====================================
Variation of g with height
(i) r
&
➔ ➔
Let r = + h
Then
If h << then
(ii) r < (We assume that the earth is a sphere of uniform density)
Consider a small mass placed at A, the spherical shell does not produce
and gravitational field on the point inside it.
&
Since we assume that the earth has uniform density
∝ r
Graphical representation
Practical data
==================================================
Altitude/ h(m) /(m)
0 9.81
1000 9.80
5 8.53
1 7.41(Parking Orbit)
3.8 0.0027 (Radius of moon orbit)
==================================================
Satellite orbit
The satellite can state in a fixed orbit with a specified radius, r, and
corresponding velocity v by F = ma
--------------- (1)
since
---------------- (2)
(2) → (1):
Note
For a satellite closed to the earth, stay at a height of about 100 –200km
Then r
Parking Orbit
The earth is rotating, at this orbit; the satellite can stay constantly
Overhead of the earth
Angular velocity of the earth
F = ma
=
h =
Note:
That means a satellite stay at a height of 36000km above the earth
Surface, can stay constantly over head when the velocity is v such that
Gravitational Potential Energy
(1) Let a point mass m is at a distance r from a sphere of mass M
- The gravitational PE is defined as the amount of work to take the mass
from ∞ to r if F, such that
Work done = w =
Note:
- The work done is negative showing energy is released during this process?
- The amount of PE at infinity is height than that at distance r
-
The PE at ∞ is zero , means that free of any attraction , the energy at ∞ must be zero
- The PE at r = work released
- The gravitation Potential : it is defined as the gravitational energy per unit mass
I.e. [V] = Jk
- The gravitational field strength (g) is defined as the -ve potential gradient.
Potential gradient =
-ve potential gradient =
Note:
The negative sign shows that the direction of g is pointing towards the centre.
************************************************************
E.g.1 Potential at the earth surface is V
Solution:
E.g.2 A double star (or binary star) consists of two stars of comparable masses M and m whose centres are d apart. They rotate together about their common centre of mass with a uniform angular velocity ω.
(a) Find the centre of mass
(b) Find the angular velocity
(c) Calculate the period of each star.
Solution:
- Taking moment about the centre of mass
Mx = m(d - x)
x =
- since The mass M revolves round the C.M.
- Period T =
E.g.3 Assume that the earth is a sphere of radius .
At what height above the earth surface has the value of g
decreased by 0.2%.
Solution:
By Binomial Theorem
E.g. 4 If the earth has a uniform density and a radius R =,
Find the percentage change in the period of a simple pendulum if it
is placed in a mine 3200m below the surface of the earth.
Solution:
&
➔
Inside the earth g ∝ r i.e.
&
E.g. 5 The moon takes 27.5 days to go around the earth once. If the orbit is
circular and has a radius 60.1 times the radius of the earth
(), Determine the value of g.
Solution:
On the earth
GM = ---------- (1)
➔ ---------- (2)
- → (1) :
E.g.6 Suppose that the earth is a solid homogeneous sphere of mass M and
radius R. Determine the gravitational potential at a point distance r
from centre, where r < R.
Solution:
When r < R
➔ ➔
When r = R
➔
E.g.7 The masses of the earth and the moon are respectively
and and they are separated by a
distance
- Ignore the motions the earth and the moon , sketch the
gravitational potential V along the line joining them.
- Find the point where the net gravitational field strength (g) is
zero.
Solution:
- Choose the earth to be the origin
V = Potential of earth + Potential of moon = +
Near the earth, V is dominated by the field of the earth and near the
moon, V is dominated by the field of the moon.
E.g.8 Some satellite, with the necessary electronic equipment inside, rises
vertically from the equator when it is fired. At a particular height
the satellite is given a horizontal momentum by firing rocket on its
surface and the satellite then turns into the required orbit.
A satellite is to be put into orbit 500km above the earth’s surface.
If it’s vertical velocity after launching is 2000 at this height.
Calculate the impulse required to put the satellite directly into
Orbit, if its mass is 50kg. (g = 10 Radius of earth =6400km)
solution:
Suppose u is the velocity required for the orbit, radius R. Then
Force on the satellite =
u = 7700
Let the impulse P required has the relation shown such that
END