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# Gravitation - Kepler and Newton revision notes and calculations.

Extracts from this document...

Introduction

Gravitation

Kepler (1571-1630) has studied for many years the records of observation on planets and summarised three laws.

Kepler’s Law

1. Each planet moves in an ellipse which has the sun at one focus
2. The line joining the sun and the moving planet sweeps out equal area in equal time
3. the square of the time of revolution of any planet (i.e. T) about the sun is proportional to the cube of the planets’ mean distance from the sun

I.e.   is a constant

Interpretation form Kepler’s laws

1. Kepler’s second law :

The area swept out in a very short time interval (Δt), neglecting the

small triangular region is A

The rate of area swept =

Hence it is a constant.

Compare this equation with the angular momentum

= Constant

This law is in fact an evidence of conservation of angular momentum.

1. Kepler’s third law

About 1666, Newton investigated the motion of the moon, and thought that it was the force of gravity to pull the moon and keep it in its orbit

NB: Time between full moon is 29.5 days but this due to the earth also moving round the sun, the moon is therefore to travel a bit longer

At the earth surface g=9.81m . This is due to the fact that we are nearer the earth centre than the moon (about 1:60)

Middle

m= mg - m

= g -    (< g )

1. At latitude θ

The body describes a circle of radius r centre

Note:

1. 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 theequator

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

Angular velocity of the earth

F = ma

=

h =

Note:

Conclusion

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

1.  Ignore the motions the earth and the moon , sketch the

gravitational potential V along the line joining them.

1. Find the point where the net gravitational field strength (g) is

zero.

Solution:

1. 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

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