For clarity we’ll use the symbol EP instead of U to denote gravitational potential energy calculated using the more accurate formula, even though the two symbols are really interchangeable. Potential energy is energy stored by doing work against any force (such as gravity) that depends only on position; therefore, gravitational potential energy EP is energy stored by doing work against the force of gravity. It can be shown (using calculus to derive the work done against gravity by changing the separation of two masses) that:
E Gm m
P = − 1 2r
where m1 and m2 are two masses separated by a displacement (or separation) r and G is the universal gravitational constant. Note that EP is always negative and approaches zero as displacement r approaches infinity (Figure 1.3.1). EP for a separation r is the work that would need to be done by a force opposed to gravity in moving the masses together, starting at ‘infinite’ separation where EP = 0 and bringing them to a separation of r (with no net change in speed).
Equivalently, EP is the work done by gravity while the masses are moved apart, starting at a separation of r to a position of ‘infinite’ separation (with no net change in speed). The gravitational potential energy does not depend on the path taken by the masses to get to their final positions; it depends only on the final separation r.
The formula isn’t affected by the choice of which mass to move, although normally we treat a large mass such as the Sun or a planet as an immoveable central body and the smaller mass as a moveable test mass. The formula seems to imply that EP approaches negative infinity as the test mass approaches the centre of a planet. However, this formula no longer applies in this form once one mass penetrates the surface of the other.
Explain that a change in gravitational potential energy is related to work done.
Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field:
E Gm m
P= − 1 2r
CheCkPoInt 1.2
1 Write down Newton’s law of universal gravitation.
2 Define weight.
3 What part of Newton’s formula for gravitational force is responsible for the inverse square law behaviour?
4 What are two names for the quantity g?
5 List three factors responsible for (real) variations in g around the Earth.
6 Outline the differences between G and g.
Worked example questIon
A piece of space junk of mass mJ drops from rest from a position of 30 000 km from the Earth’s centre. Calculate the final speed vf it attains when it reaches a height of 1000 km above the Earth’s surface. Assume that above 1000 km, air resistance is negligible.
Data: Earth’s mass mE = 5.97 × 1024 kg Earth’s radius rE = 6.37 × 106 m
Universal gravitational constant G = 6.67 × 10–11 N m2 kg–2 solutIon
Air resistance is negligible, so total mechanical energy (kinetic + potential energy) is conserved. Assume that because of the enormous mass of the Earth, its change in velocity is negligible. Use the Earth as the frame of reference. Don’t forget to convert to SI units.
Ki+ EPi = Kf + EP Cancel mJ: 1
2 mJvi2 – Gm m
rJ Ei = 12 mJvf2 – Gm m rJ E
f
Substitute: 0 – 6.67 10× –11× × f 6.67 10–
× = − ×
5 97 10 30 0 10
1 2
24
6 2
.
. v 111× ×
+ ×
5 97 10 6 37 1 00 10
24 6
.
( . . )
Rearrange, solve: vf= ×6.67 10× –11× × × −
− −
2 5 97 10 10
7 37 10
30 0
24 6 6
. . .
=9030m s−1=9 03. km s−1 Note that this result doesn’t depend on m J.
Figure 1.3.1 Plots of gravitational force (FG) and gravitational potential energy (EP) versus separation between a test mass mt and the Earth mE, starting at one Earth radius rE. The vertical FG and EP axes are not drawn to the same scale.
FG
EP –GmtmE
rE +GmtmE
rE 2rE
0 3rE 4rE 5rE
separation rE2
Isaac Newton showed that what goes up doesn’t necessarily come down. Normally, if one fires a projectile straight up, the object will decelerate until its velocity changes sign and it falls back down. However, if a projectile’s initial velocity is high enough, the 1/d 2 term in the gravity equation will cause the acceleration g to decrease with height too rapidly to bring the projectile to a stop so it will never turn back—it can ‘escape’ the planet’s gravitational field. The minimum velocity that allows this is called the escape velocity. Strictly speaking, it’s really a speed, because the initial direction of the projectile isn’t critical.
Newton treated the projectile as a cannonball (with no thrust) so that, other than the initial impulse from the cannon, the only force acting on it is gravity.
He conceived escape velocity using his force equation, and the escape velocity formula can be derived from it. However, a more modern derivation using energy is easier and similar to the previous worked example.
Let m be the mass of a projectile, M the mass of a planet, ve the initial speed and r the initial position (the planet’s radius if you are on the surface). Assume air resistance is negligible, so total mechanical energy (KE + GPE) is conserved (see in2 Physics @ Preliminary section 4.2).
Ki + EPi = Kf + EPf
The escape velocity represents the minimum limiting case where the projectile
‘just reaches infinite displacement’ with zero speed; in other words, Kf = EPf = 0.
1
2mv2 GmM 0 0
e − r = +
Rearrange, cancel m:
v GM
e= 2r
If the initial speed is greater than this, the projectile will maintain a non-zero speed even as it approaches infinite displacement. Note that the escape velocity depends only on the planet’s mass and the projectile’s starting position r but not on the projectile’s mass.
You may be puzzled that in the above derivation, the total mechanical energy (sum of KE and GPE) was exactly zero. This means that the escaping projectile has just enough (positive) KE to overcome its negative potential energy. When the mechanical energy is less than zero, there is not enough KE to overcome the GPE and the two masses are said to be gravitationally bound. When the total mechanical energy ME > 0, the KE can overcome the GPE and the two bodies are no longer bound together. This concept of binding also applies to the other three fundamental forces (including electromagnetism, which binds electrons to the nucleus of an atom).
The escape velocity from the Earth’s surface is:
2 6.67 10
m s
× × –11× ×
× 5 97 10 = − =
6 37 106. 24 11 200 1 11 2
. . km s−1
Outline Newton’s concept of escape velocity.
Explain the concept of escape velocity in terms of the:
– gravitational constant – mass and radius of the planet.
This idealised escape velocity needs to be modified when applied to real spacecraft. First, the derivation ignores air resistance in the atmosphere (hundreds of kilometres thick), which would increase the escape velocity.
Second, in a real rocket, engines produce an extra force—thrust—that can accelerate a craft to a higher altitude where the escape velocity is lower. It also ignores other sources of gravitational fields such as the Sun, Moon and planets.
The escape velocity for a projectile under the gravitational influence of more than one body is given by:
vetotal = ve12+ ve22+
where vetotal is the escape velocity for the total system and ve1, ve2 … are the escape velocities from the individual bodies within the system, calculated for the projectile using the same starting position in space.