18 - Gravitational Fields Flashcards
effect of us on the earth
we pull up the earth with the same amount of force we put on the earth - forces equal due to N3L but accelerations are different
earth is attracted to us but does not experience it as is mass is so big, a=F/m
newtons square law of gravitation
grav. force between two bodies, m and M, is proportional to mM and inversely proportional to the square of the distance between their centres
Fgrav ∝ mM/r^2
full mathematical expression of Newton’s law of universal gravitation
Fgrav = -GmM/r^2
Universal gravitational constant: G = 6.67 × 10–11 Nm2kg–2
why is gravitational force only significant when at least one of the masses involved is very large
G is a very small number
why is force have a negative sign
to indicate that the force is attractive
The Earth’s gravitational field appears to act from…
it’s centre of mass.
how does grav field strength change as distance increases from surface of earth
The Earth’s gravitational field gets weaker with distance from the Earth, this is shown by the greater separation of field lines.
new equation for grav. field strength
g=F/m and F=-GmM/r^2
> g=-GM/r^2
M > massive object causing grav. field
Define gravitational field strength at a point in space.
the force per unit mass an object feels whilst in orbit of a large mass (causing grav. field) at that point.
relationship between velocity and radius of a stable
geostationary orbit
Fcentriptal = Fgrav
mv^2/r = GMm/r^2
v=root(GM/r)
name for being closest anf furthest away from sun in an eliptical orbit
closest to sun - perihelion (perigee for moon)
furthest from son - aphelion (apogee for moon)
first law of planetary motion
the orbit of a planet is an ellipse with the sun at one focus
(eccentricity (how long and narrow) is low so motion can be modelled as circular)
second law of planetary motion
a line joining a planet and the sun sweeps out equal areas during equal intervals of time
(due to moving faster when closer to sun)
orbital period
t=d/v
T=2πr/v (time for one complete circuit)
keplers 3rd law equation
v^2=GM/r and T=2πr/v
T^2 = (4π^2/GM)r^3
T^2∝r^3
keplers third law
T^2∝r^3
The square of a planet’s orbital period is proportional to the cube of the radius of its orbit
what is a satellite
an object that orbits a larger object
examples of satellites
natural - moons, comets, asteroids
artificial - ISS, HST
what does a stable orbit depend on
distance and velocity
satellite that orbit the earth lower..
require to be faster in order to stay in orbit
v= rootGM/r
define gravitational potential
the work done per unit mass to move an object to that point from infinity
Vg=Ep/m
so units Jkg-1
GPE and height
GPE increasess as height increases as energy is required to reach the higher, doing work
GPE is defined as
being 0 at infinity as distance is too large and force of attraction is small
potential gradient
=ΔV/Δr
Vg = -GM/r
so potential gradient = -GM/r^2
potential gradient = Vg/r = g = grav field strength
g acts in opposite direction to potential gradient hence the negative sign
gravitational equipotentials
no word done by grav force when mass moved along equipotential surface
perpendicular to field lines
e.g. contour lines
potential changes more rapidly when closer to earth
how are orbits modelled as circular
the centripetal force on the satellite or planet is provided by the gravitational force of the mass
general uses of satellites
communications - satellite phones, satellite TV…
military uses - reconnaissance
scientific research - earth + universe
weather, climate - predicting + monitoring
global positioning
polar orbits
circles the poles so offers a complete view of Earth as earth rotates beneath path of orbit (new orbital path each time)
useful for global mapping and reconnaissance
low Earth orbits
short time to orbit Earth (less than 2 hours) due to short orbital radius, keplers law equation
uses include communications
geostationary orbit
remains above same point on earth while earth rotates
continuous coverage of a certain area
orbits above equator
rotates in same direction as earths rotation
has orbital period of 24 hours
uses - communications, GPS, satellite TV, weather forecasting
gravitational potential and infinity
infinity is so far that grav field strength at that point is 0, Vg is 0 at infinity (maximum) and so all values of Vg are negative as it takes external energy to move the object out of the field as all masses attract each other
if it takes 20MJ to move a unit mass to infinity, how much energy is needed to move a 2kg object to infinity?
40MJ
Vp=E/m
equation for Vg and derivation
Vg=-GM/r
using newtons square law of gravitation
and W=Fd
Vg proportionality
Vg=-GM/r
Vg is proportional to M and inversely proportional to r
where M is mass of point mass (e.g. Earth)
and r is the distance from point mass causing grav field
r is the distance…
from the centre of the planet
Vg graph
V against r:
Vg is proportional to 1/r
Vg tends to 0 as r approaches infinity
smallest value of r is equal to radius of earth and gives maximum magnitude of Vg
Vg against 1/r:
produces straight line through origin
gradient equal to -GM
change in Vg
moving towards point mass - decrease in Vg
moving away from point mass - increase in Vg
what type of quantity is gravitational potential
scalar quantity
… so total gravitational potential at any point is equal to the algebraic sum of the gravitational potentials from each mass at that point.
Vg graph (including moon)
Vg rises as you move away from earth surface, reaching a max (not 0) and falls again as distance to moon reduces
grav potential energy eqaution
E=mVg
energy = mass x grav potential
ΔE=mΔVg (m is constant)
grav potential and grav potential energy relationship
change in Vg results in change in E
so no change in Vg results in no change in E
as Vg decreases, E decreases (this energy is usually transferred into kinetic energy as object accelerates)
grav potential energy equation and derivation
Vg=-GM/r and E=mVg
so E=-GMm/r
> change in r results in change in Vg so change in E
energy needed for orbit
GPE is not the energy required to lift the satellite into orbit as it already had GPE on surface of earth. it will also need some kinetic energy in orbit
graph of gravitational force against r
F=-GMm/r^2
so graph of F against r is inverse square relationship
F ∝ 1/r^2
area under force-radius graph
work done to move mass
escape velocity
the MINIMUM velocity required for a mass to escape the gravitational field of another mass
energy required must equal the gain in gravitational potential energy
KE transferred into GPE (loss of KE = gain in GPE)
equating both prior results in v=root2GM/r
(escape velocity is same for all objects regardless of their individual masses)
