Essential idea: Similar approaches can be taken in analysing electrical and gravitational potential problems.
Nature of science:
Communication of scientific explanations: The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate, analyse and report findings to a general public used to scientific discoveries based on tangible and discernible evidence. (5.1)
Understandings:
Potential and potential energy
Potential gradient
Potential difference
Escape speed
Orbital motion, orbital speed and orbital energy
Forces and inverse-square law behaviour
Applications and skills:
Determining the potential energy of a point mass and the potential energy of a point charge
Solving problems involving potential energy
Determining the potential inside a charged sphere
Solving problems involving the speed required for an object to go into orbit around a planet and for an object to escape the gravitational field of a planet
Solving problems involving orbital energy of charged particles in circular orbital motion and masses in circular orbital motion
Solving problems involving forces on charges and masses in radial and uniform fields
Guidance:
Orbital motion of a satellite around a planet is restricted to a consideration of circular orbits (links to 6.1 and 6.2)
Both uniform and radial fields need to be considered
Students should recognize that lines of force can be two-dimensional representations of three-dimensional fields
Students should assume that the electric field everywhere between parallel plates is uniform with edge effects occurring beyond the limits of the plates
Data booklet reference:
Utilization:
The global positioning system depends on complete understanding of satellite motion
Geostationary/polar satellites
The acceleration of charged particles in particle accelerators and in many medical imaging devices depends on the presence of electric fields (see Physics option sub-topic C.4)
Aims:
Aim 2: Newton’s law of gravitation and Coulomb’s law form part of the structure known as “classical physics”. This body of knowledge has provided the methods and tools of analysis up to the advent of the theory of relativity and the quantum theory
Aim 4: the theories of gravitation and electrostatic interactions allows for a great synthesis in the description of a large number of phenomena
The gravitational potential energy of an object at a point P is equal to the work done required to take the object from infinity to the point P.
As gravitational forces are attractive, the work done required to bring an object from infinity to any point is negative. Thus, gravitational potential energy is always negative.
The gravitational potential energy of a system of two objects with mass M and m is given by
A satellite is in an orbit around the Earth. It is moved to a new orbit that is closer to the surface of the Earth. Which of the following correctly describes the changes in the gravitational potential energy and in the orbital speed of the satellite?
potential energy speed
A. increases increases
B. increases decreases
C. decreases increases
D. decreases decreases
The electric potential energy can be defined as the capacity for doing work by a change in position of the positive test charge.
Electric potential is also known as voltage. Therefore, E=V q or V=E/q
The electric potential energy is given by
were k is Coulomb’s constant, Q is the fixed charge, q is the test charge, and r is the radius.
The gravitational potential gradient of a gravitational field is given by ΔV/Δr where ΔV is the change in gravitational potential between two points and Δr is the distance between those two points.
It is the slope of a graph which plots the gravitational potential against the distance from the mass.
Gravitational potential gradient is related to the gravitational field strength (g) by g=-ΔV/Δr=GM/r^2.
Calculate the force of gravitational attraction between the Sun and the Earth. Given: mass of the Sun = 2x10^30 kg, mass of the Earth = 6x10^24 kg, distance between the center of the Sun and the center of the Earth = 1,5x10^11 m and G = 6.7x10^-11 N.m2/kg2.
Determine the orbital velocity of the Earth.