Usage Tip: This formula gives the magnitude of gravitational attraction between the masses and is exerted equally on both masses. Be careful when you use it, it may not give you the answer you require.
Usage Tip: This formula is derived from Universal Gravitation by F = ma, in this case a repre- senting g. Substituting mg for F in the universal gravitation formula and cancelling the mass of the object from F = mg gives the gravitational acceleration. The radius is required to calculate the acceleration at the surface of the object. Alternatively, the distance from the centre of the object could be used- to calculate the acceleration at a point above the surface, it would be (robject +
altitude)2. Note also that because the bottom of the equation is r2, gravitational acceleration and
therefore field strength obeys the inverse square law.
W = mg
Weight force = mass of the object × gravitational acceleration
Usage Tip: Weight force is simply the force that a gravitational field exerts on an object. As the formula shows, this force increases as the mass of the object increases so that any object in the field accelerates at a constant rate.
Ep = −
gravitational constant(6.67 × 10 ) × first mass × second mass
Potential energy = −
distance between the objects
Usage Tip: The reason this formula is negative is actually simple. When you move an object away from a gravitational field, it gains energy. A stone at the top of a building has more Ep than a stone on the ground. However, at an infinite distance from a planet, the field strength is 0 and the object doesn’t have any energy due to the gravitational field. So because Ep increases as the distance increases, and is 0 when distance is infinite, Ep takes a negative value. In this definition, it’s the work done to move an object from an infinite distance to a point in a gravitational field. Note also that the bottom of the formula is simply r, not r2. The equation is similar to Universal Gravitation but with those two key differences.
v = u + at v2 = u2 + 2as
s = ut + 1 at2
v = final velocity (m/s) u = initial velocity (m/s) a = acceleration (m/s2) s = displacement (m)
t = time (s)
Usage Tip: Select one based on the unknown required. There are 5 variables between them. Ques- tions will provide 3 variable values, so the choice of equation will be the equation with those 3 values and the variable required to solve for.
vescape = Escape velocity =
. 2 × gravitational constant(6.67 × 10−11) × mass of the planet
radius of the planet
Usage Tip: This lends itself to two key observations. Firstly, the obvious fact that the greater the mass of the planet, the higher the escape velocity, but further, that the bigger the planet, the lower the escape velocity, regardless of mass. The key reason for this is that the escape velocity depends on the distance not from the planet, but from the centre of the field. Assuming a big planet and small planet have the same mass, a rocket at the surface of the big planet will be further from the centre of the field than a rocket at the surface of the small planet. Because the rocket on the larger planet starts out further from the centre of the field, the escape velocity is lower
Fgravities = Force (in G’s) =
mg + ma mgearth
mass × acceleration due to gravity + mass × acceleration of reference frame
mass × Earth’s gravitational acceleration (9.8m/s2)
Usage Tip: G-Force is essentially the ratio of the force experienced by an object to the force it experiences at rest on Earth. This means that 2 G’s are eqivalent to twice the Earth’s gravitational acceleration (ie. 2 G’s is 19.6m/s2 ). When calculating G Force, make sure that you add on the effect of any gravitational field present- not just the acceleration of a rocket etc.
Fc = Centripetal force =
mass of the object × the object’s velocity2
radius of its circular motion
Usage Tip: Centripetal force is the force required to keep an object in circular motion at a specified
velocity. From F = ma, it can be seen that centripetal acceleration is simply v . Don’t forget in an
orbit, r is equal to orbital altitude plus the radius of the Earth, orbital radius being measured from the Earth’s centre (i.e. the centre of the field).
T = 2πr
2 × π × radius of the orbit
Usage Tip: As can be seen here, the period is just the time taken for the satellite to travel along the length of a circular orbit i.e. the circumference
v = Orbital velocity =
. gravitational constant(6.67 × 10−11 ) × mass of the planet or central body
Usage Tip: Again, orbital radius is measured to the Earth’s centre and is not the same as orbital altitude
Kepler’s Law of Periods
T 2 =
Orbital radius3 Orbital period2 =
gravitational constant(6.67 × 10 ) × mass of the planet 4 × π2
Usage Tip: This is derived by taking the orbital period formula, switching v and T , then substituting
into the orbital velocity formula and rearranging. Essentially, it states that in any system, r for any
satellite is a constant, as long as they are orbiting the same central body. Kepler’s law is used for calculating orbital radius or orbital period when the orbit of another satellite of the same system is known, or when the mass of the central body is known and the radius or period is provided
vfinal = vinitial + 2Vinitial
Final probe velocity = initial probe velocity + 2 × planet’s initial velocity
Usage Tip: This formula may not be necessary, but it’s included in the Jacaranda textbook. It’s derived by the assumption that the slingshot effect is an elastic collision. Thus two expressions are written, one equating the total momentum of the system before and after and one equating the total kinetic energy before and after, and then solving simultaneously. It probably isn’t necessary to derive it, but in order to reduce variables, the mass of the planet is expressed as Km where m is the mass of the probe and K is a value such that Km is the mass of the planet. The K’s eventually cancel out, but are necessary to derive the equation
1 − c2
. relative velocity2
Correction factor =
1 − speed of light2
This formula is not explicitly provided by itself nor is it used by itself. But it is a key part of all calculations involving relativity that are encountered in the syllabus. Essentially, the correction factor is the mathematical factor that corrects for relativistic effects. All that needs to be known is the effect relativistic speeds have and the correction factor can be easily applied. Note also that the correction factor is always a value between 0 and 1. The syllabus deals with three scenarios- time, length and mass. Time dilates, so at relativistic speeds the time taken for an event to occur in the eyes of an observer increases. So the original time taken is divided by the correction factor so that it gets bigger, giving the observed time. Conversely, if the observed time is provided, it is multiplied by the correction factor to reduce it to the value in the other frame. Length contracts, so to calculate observed length the length in the other frame is multiplied by the correction factor so that it becomes smaller. Mass dilates at relativistic speeds, so the original mass of an object is divided by the correction factor so that the observed mass is bigger. Simply put, observed time is longer, observed length is smaller, observed mass is bigger. Apply the correction factor to real figures to agree with these ideas, and you will always be correct.
Warning: The relative velocity could exceed the speed of light, for example, if both objects were heading in opposite directions at > 0.5c each. This formula is incapable of dealing with such situations. However, these situations are not considered in the HSC course.
Usage Tip: If instead of providing a specific velocity, a question provides a multiple of c, such as a
spaceship travelling at 0.6c relative to an observer, this can be substituted directly into the formula to give √1 − 0.62, because 0.6 is the ratio of spacecraft velocity to the speed of light (i.e. v ), thus
0.62 = v
Mass and Energy
E = mc2
Energy = mass of the object × speed of light2
Usage Tip: This formula only applies to the rest state. If an object is moving, add its kinetic energy to the right side of the equation