Download Gravitation powerpoint

5. Gravitation
Rotational Motion and
Astrophysics
Advanced Higher
Recap of Higher work
We have already covered the Universal
Law of Gravitation in Higher
GMm
F 2
r
Consider the Solar System
All planets in the Solar System will orbit
the Sun in an (almost) circular orbit.
We are able to apply our knowledge of
rotational motion to the behaviour of
planets.
Force of Gravitational
Attraction
GMm
F 2
r
mv 2
2
F
 mr
r
Centripetal Force
These forces act in the same direction i.e.
towards the centre of the orbit therefore…
2
GMm mv
2
F 2 
 mr
r
r
Gravitational Field Strength
Using Newton’s Law of Gravitation and
the weight formula, we can derive a
expression to work out the gravitational
field strength of any planet, star etc.
GMm
F 2
r
F  mg
GMm

mg
2
r
GM
g 2
r
Satellites in a Circular orbit
• We are able to work out how long it will
take for a satellite to make one
complete revolution of the Earth by
again equating Newton’s Law of
Gravitation and Centripetal Force
2
mv GMm
 2
r
r
GM
v
r
2r
v
T
GM 2r
r
r

 T  2r
 2
r
T
GM
GM
3
Gravitational Potential
Gravitational Potential is defined as the
work done in moving a unit mass from
infinity to a point in space.
At infinity, gravitational potential is
defined as zero
GM
V 
r
Gravitational Potential units: Jkg-1
This is not gravitational potential energy!!
Example
(a)What is meant by the gravitational
potential at a point?
(b)Calculate the gravitational potential:
(i) at the surface of the Earth
(ii) 800 km above the Earth’s surface
Solution
(a)The gravitational potential at a point is
the work that has been done in moving
a mass from infinity to that point in
space.
GM
(b)(i) V  
r
6.67 x10 11  6 x10 24
V 
6.4 x106
V  6.25 x107 Jkg 1
GM
(ii) V  
r
6.67 x10 11  6 x10 24
V 
6.4 x106  800 x103

V  5.55 x107 Jkg 1

Escape Velocity
The escape velocity is defined as the
minimum velocity required for an object
to escape the gravitational field strength
from a point in space to infinity.
For the Earth, this is approximately
11 kms-1
The potential energy of a mass is given by:
GMm
Ep  
r
By conservation of energy: Ek + Ep = 0
1
GMm
2
mve 
0
2
r
2GM
ve 
r
2
2GM
ve 
r