Gravity is an attractive force between two masses, with the magnitude of the force being proportional to their product and inversely proportional to the square of the distance between them.
Which of the following graphs correctly shows the relationship between the gravitational force, F, between two masses and the distance, r, between them?
A B C D
Newton’s law of gravitation states that: ‘Any two point masses attract each other with a gravitational force that is directly proportional to the product of their masses and inversely proportional to the square of their separation.’
Inverse square law: If the distance is doubled the , lines spread out over four times its surface area, so their concentration is reduced by a quarter.
So the gravitational field for a spherical mass of radius 𝑅 can be represented as a radial field. The arrows always point towards the centre of a mass since gravitational force is always attractive.
The field close to the surface of a spherical mass is approximately uniform. Hence the lines are parallel, equidistant and perpendicular to the surface.
Gravitational field strength is a vector quantity in the direction of the gravitational force.
Kepler’s 3rd law: Kepler showed that, for a planet orbiting the Sun, the relationship between the orbitaltimeperiod 𝑻 and the orbitalradius 𝒓 is given by: 𝑻^𝟐 ∝ 𝒓^3
A satellite following a synchronous orbit has a time period equal to the rotational period of the planet being orbited.
A satellite following a synchronous orbit around the Earth is described as having a geosynchronous orbit.
A communication satellite requires a special type of geosynchronous orbit, e.g. a geostationary orbit.
A satellite with a geostationary orbit around a planet will remain vertically above the same point on the equator of that planet at alltimes.
The gravitational potential 𝑽 at a point is the workdone ∆�� per unit mass 𝒎 (or the change in potential energy ∆𝑬𝒑 per unit mass 𝒎) to move a mass 𝒎 from infinity to that point.
work done ∆𝑊 in moving a 𝑚 from a from infinity to a point at a distance 𝑟 from the centre of the planet of mass 𝑀 is ∆𝑾 =- GMm/r
So the gravitational potential 𝑉 at a distance 𝑟 from the centre of a planet of mass 𝑀: V= - GM/r Hence 𝑉 is inverselyproportional to 𝑟. 𝑉 is a scalar quantity.
An object’s escape velocity is the minimum velocity required at the surface of a planet in order to escape the gravitational pull of the planet usingitsownkinetic energy (i.e. no further propulsion).
The total energy of a satellite in orbit is the sum of its gravitational potential energy and its kinetic energy: 𝑬total= - GMm/2r
For a satellite in orbit: Ek= +GMm/2r
For a satellite in orbit: Ep= - GMm/r
What is Newton’s law of gravitational force?
Any two point masses that attract each other with a gravitaional force that is directly proportional to the product of their masses and inversely proportional to the square of their separation
What are the units for gravitational field strength?
Nkg^-1
What is meant by the inverse square law?
If the distance is doubled the , lines spread out over four times its surface area, so their concentration is reduced by a quarter.
What is Kepler’s 3rd law?
Kepler showed that, for a planet orbiting the Sun, the relationship between the orbital time period 𝑻 and the orbital radius 𝒓 is given by: T^2∝ r^3
What is the equation that relates orbital time period T and orbital radius r?
T2=(4π2/GM)r3
What is the equation for orbital speed v?
v2=GM/r
What is meant by a ‘Geostationary orbit’ and how is this achieved?
A satellite with a geostationary orbit around a planet will remain vertically above the same point on the equator of that planet at all times.
What is a satellite in a ‘Geostationary orbit’ used for?
Telecommunications transmissions. Satellite television broadcasting.
What are the advantages of a ‘Geostationary orbit’?
Receiving dishes can have a fixed position pointing to the same spot in the sky and can maintain continuous contact with the satellite.
What are the disadvantages of a ‘Geostationary orbit’?
Each satellite communicates with a restricted area of the Earth’s surface.
What is the definition of gravitational potential?
The gravitational potential 𝑽 at a point is the work done ∆𝑾 per unit mass 𝒎 to move a mass m from infinity to that point.
What is the equation for gravitational potential?
V=−(GM/r)
What are the units for gravitational potential?
Jkg^-1
• 𝑽 = 𝟎 at infinity.
• 𝑉 = ∆𝑤/𝑚 : Work is done per unit mass as a body moves from infinity towards a point in a field.
• Work is done by the field, so potential values become more negative as ‘r’ decreases.
What is the area under a graph of g against r?
∆𝑉 the gravitational potential
What is an equipotential?
An equipotential in a gravitational field is a surface (or line in a two-dimensional context) on which the gravitational potential is constant. This means that at every point on an equipotential surface, a mass would have the same potential energy due to the gravitational field.
Moving a mass perpendicularly across a field line does not alter its potential energy.
If a mass is moved in the direction of the field lines the mass’ potential energy is reduced.
A greater density of field lines represents a stronger field.
An object that has escaped a gravitational field has zero potential energy.
The total energy of a satellite orbiting the Earth has a negative value.
The gravitational potential gradient at a point has the same numerical value as the gravitational field strength at that point.