Gravitational fields Flashcards
Explain what is meant by the gravitational potential at a point in a gravitational field
work done against the field per unit mass when moved from infinity to the point
Use the following data to calculate the gravitational potential at the surface of the Moon.
mass of Earth = 81 × mass of Moon
radius of Earth = 3.7 × radius of Moon
gravitational potential at surface of the Earth = –63 MJ kg–1
-2.88MJ/kg
Sketch a graph to indicate how the gravitational potential varies with
distance along a line outwards from the surface of the Earth to the surface of the Moon.
Curving line upwards from limiting value at surface of earth to point close to 0 near the surface of moon then falls slightly towards it
State Newton’s law of gravitation
force of attraction between two point masses is proportional to product of masses and inversely proportional to square of distance between them
In 1798 Cavendish investigated Newton’s law by measuring the gravitational force between
two unequal uniform lead spheres. The radius of the larger sphere was 100 mm and that
of the smaller sphere was 25 mm. density of lead = 11.3 × 103
kg m–3
(i) The mass of the smaller sphere was 0.74 kg. Show that the mass of the larger
sphere was about 47 kg.
(ii) Calculate the gravitational force between the spheres when their surfaces were in
contact.
- 3kg
1. 5 × 10−7 (N)
Modifications, such as increasing the size of each sphere to produce a greater force
between them, were considered in order to improve the accuracy of Cavendish’s
experiment. Describe and explain the effect on the calculations in part (b) of doubling the
radius of both spheres.
for the spheres, mass ∝ volume
mass of either sphere would be 8 × greater
this would make the force 64 × greater
but separation would be doubled causing force to be 4 × smaller
net effect would be to make the force (64/4) = 16 × greater
Artificial satellites are used to monitor weather conditions on Earth, for surveillance
and for communications. Such satellites may be placed in a geo-synchronous orbit or in a
low polar orbit.
Describe the properties of the geo-synchronous orbit and the advantages it offers when a
satellite is used for communications.
orbits (westwards) over Equator (1)
maintains a fixed position relative to surface of Earth (1)
period is 24 hrs (1 day) or same as for Earth’s rotation (1)
offers uninterrupted communication between transmitter and receiver (1)
steerable dish not necessary (1)
A satellite of mass m travels at angular speed ω in a circular orbit at a height h above the
surface of a planet of mass M and radius R.
(i) Using these symbols, give an equation that relates the gravitational force on the
satellite to the centripetal force.
(iii) Explain why the period of a satellite in orbit around the Earth cannot be less than 85
minutes. Your answer should include a calculation to justify this value.
mass of the Earth = 6.00 × 1024 kg
kg radius of the Earth = 6.40 × 106 m
mw^2(R+h) = GMm/(R+h)^2
GMm/(R+h)^2 = (2pi/T)^2
there t^2 =
limiting case is orbit at zero height i.e. h = 0
T = 5090 s (1) (= 85 min)
Describe and explain what happens to the speed of a satellite when it moves to an orbit
that is closer to the Earth.
speed increases
loses potential energy but gains kinetic energy
State two features of a geo-synchronous orbit.
period is 24 hours = Earth’s rotation
remains in fixed position relative to surface of Earth (1)
equatorial orbit (1)
same angular speed as Earth (or equatorial surface
The mass of the Earth 6.00 × 1024 kg and its mean radius is 6.40 × 106
m.
(i) Find the radius of a geo-synchronous orbit
(ii) Calculate the increase in potential energy of a satellite of 750 kg when it is raised
from the Earth’s surface into a geo-synchronous orbit.
- 23 × 10^7m
3. 98 ×1010 J
Satellites in orbits nearer the Earth than geo-synchronous satellites may be used in the
future to track road vehicles.
(i) State and explain one reason why geo-synchronous satellites would not be suitable
for such a purpose.
signal would be too weak at large distance (1)
(or large aerial needed to detect/transmit signal, or any other
acceptable reason)
the signal spreads out more the further it travels
Satellites in orbits nearer the Earth than geo-synchronous satellites may be used in the
future to track road vehicles Give two points you would make in arguing for or against tracking road vehicles.
Explain your answers.
for: road pricing would reduce congestion
stolen vehicles can be tracked and recovered
uninsured/unlicensed vehicles can be apprehended
against: road pricing would increase cost of motoring
possibility of state surveillance/invasion of privacy
The microwave signal is received 68 ms after it was transmitted from the satellite. mass of the Earth = 6.0 × 1024 kg
mean radius of the Earth = 6400 km
Calculate the height of the satellite, the strength of the gravitational field , the speed, the time period
- 0(4) × 10^7m
- 56 N kg–1
- 9 × 103m s–1
- 3(5) × 104s
Explain why no work is done by the gravitational force that acts on the moon to keep in orbit around th earth
work = force * distance moved in direction of force
Force is perpendicular to displacement
No movement in the direction of force hence no work
Kinetic and potential energy is constant
Give an example of a situation where a body travels at constant speed but experiences a continuous acceleration
Any example of circular motion
Give an example of a situation where a body experiences a maximum acceleration when its speed is zero
Any example of super harmonic motion at maximum displacement
Show that the time taken for one orbit around the sun squared is proportional to the radius of the orbit cubed
mw^2 = GMm/R^3
T = 2pi/w
(2pi/T)^2 = GM/R^3
hence T^2 is proportional to R^3
The Earth’s orbit is of radius 1.510^11
The mean radius of Mercury’s orbit is 5.7910^10
Neptune orbits the sun once every 165 Earth years
(i) calculate the length of Mercury’s year
(ii) calculate the ratio: distance from Sun to Neptune/distance from Sun to Earth
- 5 days
30. 1
Describe the properties of the geo-synchronous orbit and the advantages it offers when a satellite is used for communications
orbits westwards over equator
maintains a fixed position relative to surface of Earth
period is 24 hrs
steerable dish not necessary
offers uninterrupted communications between transmitter and receiver
State what is meant by gravitational field strength at a point in a gravitational field and state whether it is a scalar or vector quantity
force per unit mass
vector
Draw a graph of gravitational potential against radius
curve of 1/x + c
Show that T does not depend on its own mass
T^2 = 4*pi^2**r^3/GM
Orbital period = 7.15 earth days
Radius = 1.07*10^6
Calculate the angular speed, centripetal acceleration and mass of the planet
- 02 * 10^-15 rad/s
- 111m/s^2
- 9*10^27
Gravitational force _____ as r increases because
decreases as is proportional to r^-2
There is an attractive force between all masses: F =
GMm/r^2 (G is the universal constant of gravitation, r is the distance between M and m)
Gravitational field =
force field round a mass
A small mass is attracted to large body and follows a
field line
Radial field
field lines like wheel spokes always directed to the centre of a larger mass
The magnitude of gravity in a radial field decreases as
distance from the massive body increases
Uniform field =
gravitational field strength is the same in magnitude and direction throughout
Force field =
a region where an object experiences a force
Gravitational field lines show that
show the force that a small test mass experiences in such a region
Gravitational field strength =
force per unit mass (N/kg) that an object experiences in a field: g=F/m
In a radial gravitational field around mass M: gravitational field strength, g =
GM/r^2
Gravitational potential is
the work done in bringing an object with unit mass from infinity to a point in a gravitational field (unit J/kg): V = -GM/R
Gravitational potential difference is
the difference in gravitational potentials of two points in a gravitational field
gravitational potential is a property of a particular point in the gravitational field and does not depend on
the object in the field
V and g can be related by: gravitational field strength, g =
-deltaV/deltar
The work done in moving an object of mass m between two points in a gravitational field is =
deltaW = mDeltaV (V is scalar)
The potential gradient at a point in a gravitational field is
the change of potential per metre at that point
potential gradient =
deltaV/deltaR
The orbiting object of mass m is in
circular motion
F =
2
F=ma
F=GMm/r^2
a =
3
a=v^2/r
a=w^2r
a=GM/r^2
Period, T =
2pi/w
A lower orbit (smaller r) has ____ potential energy and ____kinetic energy than a higher orbit (bigger r)
less
more
Geosynchronous orbits have a period of
24 hours = one day
deltaW = … = …
deltaW = FdeltaR = GMm/r^2 delta r
Kepler’s third law =
t^2 is proportional to r^3
Equipotentials are
lines of constant potential
Gravitational potential energy is
the energy of an object due to its position in a gravitational field
The position for zero gpe is
At the surface, the gpe is
infinity
negative so it needs to do work to escape from the field completely
from the Earth’s surface, the equipotentials for equal increases of potential are spaced…
further apart
However near the surface, the equipotentials are
horizontal because the gravitational field over a small region is uniform.
Newton’s law of gravitation assumes that
the gravitational force between any two point objects is: always an attractive force, proportional to the mass of each object, proportional to r^-2 where r is their distance apart
To move m a small distance r in the opposite direction to the gravitational force F on it, its gravitational potential energy must be
increased: by an equal and opposite force F acting through the distance r, by an amount of energy equal to the work done by F, W=Fr
gravitational field strength of the Earth falls with _____ distance from the Earth.
increasing
However over small distances which are much less than the Earth’s radius, the change of gravitational field strength…
is insignificant. For example, the measured value of g has the same magnitude and direction 100m above the earth as it is on the surface. In theory, g is smaller higher up, but the difference is too small to be noticeable. Only over distances which are small compared with the Earth’s radius can the Earth’s field be considered uniform
Diagram of gravitational field strength against distance of radius
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