Gravitation Flashcards
the assumptions made by Newton
- all the mass of the Earth may be considered to be concentrated at the centre of the Earth.
- the gravitational attraction of the Earth is responsible for the moon’s circular motion round the Earth. Observed central acceleration can be calculated from measurements of the moon’s motion.
how to work out gdilute
1/60 squared x 9.8
Re/rm = 1/60
central acceleration of moon equation
a=v squared / r OR a = 4pi r squared / T squared
velocity of moon equation
v= 2 pi r / T
Newtons hypothesis on gravitation
inverse square law
acceleration due to gravity would quarter if the distance from the centre of the Earth doubles
compared with the diluted gravity at the radius of the moon’s orbit according to hypothesis
1/60 squared x 9.8
universal constant of gravitation
G = 6.67x10 ^-11 Nm^2kg^-2
Inverse square law of gravitation
F = GMm / r^2
Kepler had already shown r^3 / T^2 = constant and since Ms =constant…
G must be constant for all planets in the solar system
although we assume circular orbits, what are orbits actually?
elliptical
gravitational force is very weak compared to what?
electromagnetic force
how to work out the mass of the Earth
two equation: F=GMm/r^2 and F=W=mg
mg=GMem/Re^2
Me=gRE^2 / G
when we consider force and energy changes on a large scale we have to take into account…
the variation of force with distance
definition of gravitational field strength at a point
the force experienced per unit mass in a gravitational field.
g=F/M
G AND F ARE VECTORS
field lines and equipotential lines
equipotential lines are always at right angles to field lines
from far away, any gravitational field will look like a point mass
lines do not touch at the centre
equation for g
g=GM/r^2
variation of g on Earth
the density of Earth is not uniform
greatest value at sea level is found at the poles and smallest is at the equator (caused by rotation of Earth)
why is g smallest at equator
Masses at the equator experience the maximum spin of the earth. They are in circular motion with a period of 24hrs and radius 6400km. Part of mass’s weight used to supply central force due to the circular motion.
what has caused the flattening at the poles?
the centrifuge effect on the liquid Earth as it cools.
locally, g varies depending on…
underlying rocks and sediments
what is the central force required to keep satellites in orbit provided by?
the force of gravity
a satellite orbiting Earth at radius r, has an orbit period of…
T=2PI ROOT R^3/GM
Kepler’s 3rd Law of planetary motion
the two tides per day that we observe are caused by…
the unequal attractions of the moon (and sun) for masses at different sides of the Earth.
The rotation of Earth and position of moon also has an effect.
tides acting in phase
spring tides (strong)
tides acting out of phase
neap tides (less strong)
astronomical units
mean distance between the Earth and Sun - 1.5x10^11 m
light years
the distance light can travel in a year - 9.46x10^15m
gravitational potential
work done by external forces in moving a unit mass from infinity to that point
gravitational potential equation
V=-Gm/r *units J/Kg
gravitational potential energy
work done in moving mass from infinity to a point in the gravitational field.
gravitational potential energy equation
Ep=-GMm/r
total energy of satellite equation
Etotal = -GMm/2r
escape velocitry
the minimum velocity the mass must have which would allow it to escape the gravitational field.
escape velocity equation
Ve = ROOT 2GM/r
escape velocity on Earth
approx 11km/s
atmospheric consequences
small no of H2 molecules have velocity greater than ve so there is a loss of hydrogen in the Earth’s atmosphere.
why does the moon have no atmosphere?
ve is 2.4km/s so gaseous molecules have enough energy to escape from the moon
