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a gravitational field is
a region where any other mass will experience a force

the gravitational field strength , g , at a point us defined as the
 force that acts on a unit mass placed at that point
 g = F/m
 units Nkg^{1}
 vector quantity

a gravitational field can be represented by
 sketching field lines around a mass
 the field lines show the direction of force that acts on another mass

near the earths surface the field lines are approx
parallel to each other and perpendicular to the surface of the earth , the field is said to be uniform  the gravitational field strength is the same magnitude and direction throughout the field

on a larger scale the field lines around a spherical mass
spread out radially

force of gravity on a small mass near a much larger spherical mass is always directed to
the centre of the larger mass

the magnitude of g in a radial field .... with increased distance from the massive body
decreases

over small distances which are much less than the earths radius the change in gravitational field strength is
insignificant so the field is considered to be uniform

for a radial field such as that due to a spherical mass , M , the force that acts on a small mass , m , placed at distance , r, is given by Newtons law as
F = GMm/r^{2}

g = F/m therefor g also =
 g = GMm/mr^{2}
 so
 g = GM/r^{r}

inside the earth as r becomes smaller the mass becomes smaller so g
decreases

draw a graph to show how g changes with r

as a mass , m , moves closer to the earth its potential energy changes , usually
Ep = mgΔh

however for large distances
g isn't constant

in order to be able to calculate energy changes in a gravitational field we define a a quantity called
the gravitational potential

the gravitational potential at a point in a gravitational field is defined as
the work done in bringing a unit mass from infinity to that point

the gravitational potential at a point X from the surface of a planet is
equal to the work done in bringing a unit mass from infinity to point X

work done =
force x distance moved in direction of force

the gravitational force on a unit mass at distance r from a mass , M , is F_{}= GMm/r^{2} so the work done in moving the mass through a distance Δr
ΔrGMm/r^{2} . the negative sign shows that the force F acts in the opposite direction to displacement

gravitation potential =
W/m

draw a graph to show how gravitational potential varies with distance from the earth
y axis should be force on a 1kg mass due to Earths gravitational field for the purpose of the next question only

the total work done in moving from infinity to a distance r is
 the area under the graph between r and infinity
 each strip has an area FΔr =  GMmr/r^{2}_{ =}
 FΔr =  GMm/r
 so V = FΔr/m = GM/r

gravitational potential is a ... qauntity
scalar

units of gravitational potential are
JKg^{1}

the negative sign indicates that
this is abound system : energy would have to be transferred to the unit mass in order to pull it away

http://www.vvc.edu/academic/chemistry/Unit%209R%20%20Carboxylic%20Acids.pdf
for chemistry


the gradient of a gravitational potential graph vs distance is
g

the work done in moving a mass in a gravitational field corresponds to
change in gravitational potential energy

change in gravitational potential energy can be thought of as
change in potential energy per unit mass

to calculate the potential energy we need to
 multiply the gravitational potential by the mass that's in the field
 E_{p}=mV

the magnitude of the gravitational potential at a point is
the amount of energy needed to move a unit mass from a point to infinity

hence the magnitude of the gravitational potential is
the amount of energy required to pull the mass completely out of the gravitational field

this energy could come from
the initial kinetic energy

this means we can calculate how fast a mass has to be thrown vertically up so that is never comes down again this is known as
escape velocity

if a mass has enough energy to escape then neglecting work done against air resistance
ΔE_{k} =
 ΔE_{p}
 0.5mv^{2}=mV (V = GM/r)
 0.5mv^{2}=mGM/r
 so the escape velocity is √mGm/r0.5m
 √2GM/r

gravitational field strength =
force per unit mass

gravitational potential =
energy per unit mass

in any gravitational field the greater the field strength the .... the gravitational potential changes with distance
more

the rate of change of gravitational potential with distance is equal to
the field strength g = V/r

what are the units of equipotentials
MJkg^{1}

equipotentials are similar to contour lines on a map they
 connect points where a mass would have the same gravitational potential energy

as a mass moves in a gravitational field its .... changes due to the force of gravity
velocity

the exact shape of the path that it follows depends on the initial velocity as well as
the strength of the field

if we consider faster moving objects like a spacecraft the general shape of the path
is a hyperbola if v > escape velocity

for an object above the earth F =
gravitational force

at a particular speed the spacecraft can go into circular orbit around the planet . this happens when
the gravitational force (force between two masses) equals the centripetal force necessary to make the spacecraft move in a circular path

what is the equation for the velocity at which the spacecraft can go into circular orbit
 F=GMm/r^{2} F=mv^{2}/r (where r is the distance between orbit and centre earth)
 GMm/r^{2} = mv^{2}/r
 GM/r = V^{2}
 √GM/r = v

the orbital speed depends on
the radius of the orbit and the mass of the planet but not the mass of the satellite

for artificial satellites orbiting the earth those in a lower orbit have higher orbital speeds because
v = √GM/r and at lower orbits r is less

polar satellites
 close to earth
 have an orbital period of 90 minutes
 have a high orbital speed
 are weather satellites

GPS
 global positioning satellite
 orbit between polar and geostationary satellites
 need at least 3 satellites to find your position

geostationary satellites can also be called
geosynchronous satellites

geostationary satellites
 have an orbital period of 24 hours
 have a high equatorial orbit
 low orbital speed
 are satellites used for relaying TV pictures and therefor have to be in a fixed position relative to earth so the domestic satellite doesn't have to track the satellite across the sky

calculate the radius of a ... orbit
 F=GMm/r^{2}
 F = mv^{2}/r
 GMm/r^{2} = mv^{2}/r
 GM/r = v^{2}
 time = distance /speed
 T = 2pir/v
 so v = 2pir/T
 GM/v^{2} = r
 GM/(2pir/T)^{2} = r
 GMT^{2}/4pi^{2}r^{2}=r
 GMT^{2}/4pi^{2}=r^{3}
 where r is the distance from centre of earth so to find out how far above the earths surface the orbit is calculate r  radius of earth

the final formula on the previous page tells us that
T^{2} is proportional to r^{3} as G , M and pi are constants

T^{2} for earth

r^{3 }for earth
would be the same as
 T^{2} for planet y
 
 r^{3} for planet y

the total energy of a satellite is the
sum of its potential energy and its kinetic energy

if the satellite is moving at a velocity v in a circular orbit of radius r around the earth then
 E_{total }= E_{K} + E_{p }= 0.5mv^{2} + mV
 this equals 0.5mv^{2}  GMm/r

the moons orbit around the earth is assumed to be
circular

work done =
force x distance moved in the direction of the force

in circular motion force and direction are perpendicular so there's no
travel in the direction of the force so the work done is zero

at point A why doesn't the spacecraft have acceleration when the thrust motor is switched off
the force of attraction between the spacecraft and earth is equal and opposite to the force of attraction between the spacecraft and the moon so there's no resultant force and hence no acceleration

at point B without any motor thrust the speed of the spacecraft is constant but there's acceleration this is because
the spacecraft is in circular orbit around the earth so there's always a change in direction and hence velocity and hence theres acceleration . there's no change in speed because circular orbits are in uniform circular motion

at C the velocity is 6000ms^{1} it moves through an angle of 1/100 rad without changing speed this means that
thrust must be perpendicular to the direction of motion

how long can the the thrust be maintained for at C if the if the thrust is 12000N
 F = Δmv/t
 t = Δmv/F
 t = 2000x 60000x1/100
 
 12000
 t = 10 seconds

the spacecraft at D must be accelerating towards the centre of the earth if the thrust isn't working because
the only force acting on the spacecraft is the gravitational force of attraction between the earth and the spacecraft and this force is the centripetal force for the circular orbit

what is the magnitude change in momentum of the spacecraft over 100s
 F = Δmv/t
 Δmv = Ft
 F = GMm/r^{2} x t
 4x10^{14}x2000/(1000000x10^{3})^{2 }x 100 = 800Ns ^{}

why can we ignore the mass of the moon in our force calculations because
the mass of the earth is much bigger than the mass of the moon . even if they were the same mass the moon is 3 times further away than the earth so g is 9 times smaller g is proportional to 1/r^{2} . this means the effect of the moon is negligible

the thrust motors aren't in use between c and e calculate the change in gravitational potential energy
 E_{P} = mV
 mV  mV
 V = GM/r
 (GMm/r  GMm/r)
 (2000x4x10^{14}/2 x10^{8}) = 3.6x10^{9}J
 the potential energy increases between E and C because the gravitational potential is less at E hence the gravitational potential energy is less at E

the change in gravitational potential energy would be the same even if the motors were used because
the velocity is still perpendicular to thrust so ne extra work done so no change in gravitational potential and hence gravitational potential energy

at C what is the kinetic energy
0.5mv^{2} = 0.5 x 2000x6000^{2} = 3.6x10^{10}J

what is the kinetic energy at E
3.6 x 10^{10 } 3.6 x 10^{9} = 2.34 x 10^{10 }J

what would be the kinetic energy is E was in permanent orbit around the earth
 when in circular orbit v^{2} = GM/r
 so E_{k} = 0.5mGM/r
 0.5 x 2000 x 4x10^{14} / 20000x10^{3} = 2x10^{10}J

if v> escape velocity path is
hyperbola

v < escape velocity
path is elliptical

v^{2} = GM/r
circular orbit

at some point between the earth and moon g on an object = 0
 because GMm/r^{2 }= GMm/r^{2}^{}
 ^{where} GMm/r^{2}^{ is equal to the gravitational force of planet }

newtons law of gravitation
the gravitational force between two point masses is proportional to product of the masses and inversely proportional to the square of the distance between them

Ep = mV change in m
has the same effect on Ep but no effect on V

when using V =  GM/r
ignore the negative



centripetal acceleration =
g

