Gravitational Potential Energy Near Earth for HSC Physics

You lift a 5 kg backpack from the floor onto a 1.2 m bench. Has the backpack “gained 59 J of gravitational potential energy”, or does that answer depend on where you decide zero energy is?

Predict before reading on. If you call the floor zero, the backpack has positive gravitational potential energy on the bench. But if you call the bench zero, its gravitational potential energy there is zero. So which description is correct?

Both are. The actual value of gravitational potential energy depends on your chosen reference level. The physically important quantity is usually the change in gravitational potential energy:

\[
\Delta U = mg\Delta h
\]

That change does not depend on where you put zero.

01Start with the physical picture

Imagine holding a textbook above the floor. If you release it, gravity can speed it up as it falls. Raise the book higher, and gravity has more distance over which it can do work.

That is the useful idea behind gravitational potential energy near Earth’s surface. Separating an object vertically from a lower position gives the object-Earth system the capacity to transfer energy as gravity pulls the object back down.

A simple way to picture this is a staircase. Moving sideways along one step does not change your height. Climbing to the next step does. Near Earth’s surface, gravitational potential energy follows the vertical change, not the distance you travelled to get there.

This model is useful, but slightly simplified. Gravitational potential energy actually belongs to the gravitational interaction between the object and Earth, not to the object alone. In HSC problems near Earth’s surface, saying that “the object gains gravitational potential energy” is convenient shorthand.

02The equation that matters

For motion near Earth’s surface, where the gravitational field strength can be treated as approximately constant,

\[
\Delta U = mg\Delta h
\]

where:

• \(\Delta U\) is the change in gravitational potential energy, measured in joules (J)
• \(m\) is the object’s mass, measured in kilograms (kg)
• \(g\) is the magnitude of the gravitational field strength, approximately \(9.8\ \text{m s}^{-2}\) near Earth’s surface
• \(\Delta h\) is the change in vertical height, measured in metres (m)

The height change is

\[
\Delta h = h_f-h_i
\]

where \(h_i\) is the initial height and \(h_f\) is the final height relative to the same reference level.

This gives an important sign rule:

The sign is telling you what happened to the energy of the gravitational interaction. Moving upward increases gravitational potential energy. Moving downward decreases it.

03Why the reference level can be chosen

Suppose a 2 kg object is sitting on a shelf 3 m above the floor.

If the floor is \(h=0\),

\[
U = mgh = (2)(9.8)(3)=58.8\ \text{J}
\]

Now imagine choosing the shelf itself as \(h=0\). The same object now has

\[
U=0\ \text{J}
\]

Nothing physical changed. The object did not suddenly lose 58.8 J because you changed your coordinate system.

This sounds strange until you compare it with altitude. A person’s height above sea level and height above the ground outside their house can have different numerical values. The person’s actual location has not changed. Only the reference has.

Gravitational potential energy works similarly.

The analogy has a limit. Height is a position coordinate, while gravitational potential energy describes an interaction and depends on mass as well as position. Still, the reference-level idea is the same.

What stays unchanged?

Suppose the object moves from a shelf 3 m above the floor to a table 1 m above the floor.

Using the floor as the reference,

\[
U_i=(2)(9.8)(3)=58.8\ \text{J}
\]

and

\[
U_f=(2)(9.8)(1)=19.6\ \text{J}
\]

so

\[
\Delta U=U_f-U_i=19.6-58.8=-39.2\ \text{J}
\]

Now choose the table as the reference instead. The initial shelf is 2 m above the reference, while the final position is at \(0\ \text{m}\):

\[
U_i=(2)(9.8)(2)=39.2\ \text{J}
\]

\[
U_f=0\ \text{J}
\]

and again,

\[
\Delta U=0-39.2=-39.2\ \text{J}
\]

The individual values of \(U\) changed. The change \(\Delta U\) did not.

That is why HSC problems are usually much more interested in \(\Delta U\) than in an isolated value of \(U\).

04How to choose a useful reference level

Mathematically, you can place \(U=0\) wherever you like. Practically, some choices make a problem much easier to read.

Choose a reference that makes the important heights simple.

Good choices often include:

• the ground
• a floor or tabletop
• the lowest point of the motion
• the object’s initial position
• any clearly defined horizontal level used repeatedly in the problem

For example, if a ball is thrown upward from a balcony and you only care about energy changes from the balcony, setting the balcony to \(h=0\) can remove unnecessary numbers.

The choice is not about finding the “correct” zero. There is no unique zero for near-Earth gravitational potential energy. The requirement is consistency.

Once you choose a reference, every height in that calculation must be measured from that same level.

06Path length is not height change

Suppose you carry a box up a long ramp that rises \(2.0\ \text{m}\). The ramp itself is \(8.0\ \text{m}\) long.

Which distance belongs in

\[
\Delta U=mg\Delta h
\]

Predict before continuing.

It is \(2.0\ \text{m}\), not \(8.0\ \text{m}\).

The equation contains \(\Delta h\), the vertical displacement, rather than the distance travelled along the path.

A \(10\ \text{kg}\) box raised \(2.0\ \text{m}\) gains

\[
\Delta U=(10)(9.8)(2.0)=196\ \text{J}
\]

whether it travels:

• straight upward,
• along a ramp,
• up a staircase, or
• along some unnecessarily dramatic zigzag route.

Other forces might do different amounts of work on different paths, especially if friction is involved. But the change in gravitational potential energy depends only on the starting and finishing heights.

This is one consequence of gravity being a conservative force.

08The tempting misconception: gravitational potential energy cannot be negative

A student sees

\[
U=mgh
\]

and reasons that mass and \(g\) are positive, so gravitational potential energy must also be positive.

That feels sensible until you remember that \(h\) is measured relative to a chosen zero.

If you define the top of a \(4\ \text{m}\) shaft as \(h=0\), then a point \(3\ \text{m}\) below it has

\[
h=-3\ \text{m}
\]

A \(2.0\ \text{kg}\) object there would have

\[
U=(2.0)(9.8)(-3)=-58.8\ \text{J}
\]

There is nothing physically wrong with this result.

A negative value does not mean the system contains an impossible amount of energy. It means the object is below the level you chose to call zero.

What matters physically is how energy changes between positions.

09Do not confuse \(U\) with \(\Delta U\)

This distinction causes a lot of avoidable errors.

The expression

\[
U=mgh
\]

gives gravitational potential energy relative to a chosen reference level, provided \(h\) is the object’s height relative to that level.

The expression

\[
\Delta U=mg\Delta h
\]

gives the change between two positions.

For example, suppose a ball moves from \(h=8\ \text{m}\) to \(h=3\ \text{m}\).

Its final height is still positive, so \(U_f=m g(3)\) is positive if the reference is \(h=0\).

But its change in height is

\[
\Delta h=3-8=-5\ \text{m}
\]

so

\[
\Delta U=mg(-5)
\]

is negative.

The ball can therefore have positive gravitational potential energy while its gravitational potential energy is decreasing.

Those statements are not contradictory.

10A reliable HSC method

For most near-Earth gravitational potential energy calculations:

1. Choose a clear reference level if one has not already been given.
2. Write the initial and final vertical heights relative to that reference.
3. Calculate \(\Delta h=h_f-h_i\).
4. Use \(\Delta U=mg\Delta h\).
5. Keep the sign of \(\Delta h\). Do not replace it with a positive distance.
6. State what the sign means physically.

This method is safer than trying to decide in advance whether to “add” or “subtract” energy.

11When the equation is an approximation

The equation

\[
\Delta U=mg\Delta h
\]

treats \(g\) as constant.

That is an excellent approximation for ordinary height changes near Earth’s surface because those height changes are tiny compared with Earth’s radius.

It is not the general equation for gravitational potential energy everywhere in space.

As an object moves very far from Earth, gravitational field strength decreases noticeably with distance. For those situations, gravitational potential energy must be described using the universal gravitational model,

\[
U=-\frac{GMm}{r}
\]

where \(G\) is the universal gravitational constant, \(M\) is Earth’s mass, \(m\) is the object’s mass, and \(r\) is the distance between the centres of Earth and the object.

Near Earth’s surface, however, \(\Delta U=mg\Delta h\) is the useful local model.

12Questions and solutions

Question 1

A \(3.5\ \text{kg}\) bag is lifted from a chair \(0.50\ \text{m}\) above the floor to a shelf \(1.70\ \text{m}\) above the floor. Calculate the change in gravitational potential energy of the bag-Earth system.

Solution 1

The gravitational potential energy increases by approximately \(41\ \text{J}\).

The important quantity is the change in vertical height:

\[
\Delta h=h_f-h_i=1.70-0.50=1.20\ \text{m}
\]

Using

\[
\Delta U=mg\Delta h
\]

gives

\[
\begin{aligned}
\Delta U&=(3.5)(9.8)(1.20)\\
&=41.16\ \text{J}
\end{aligned}
\]

Therefore,

\[
\boxed{\Delta U\approx +41\ \text{J}}
\]

The positive sign shows that the bag finishes higher than it starts. A common error is to use \(1.70\ \text{m}\) as \(\Delta h\), but that is the final height above the floor, not the change in height.

Question 2

A \(4.0\ \text{kg}\) crate is moved from a loading platform \(2.5\ \text{m}\) above the ground down to a trolley \(0.80\ \text{m}\) above the ground.

Calculate \(\Delta U\). Then state whether choosing the loading platform as \(h=0\) would change your result.

Solution 2

The change in gravitational potential energy is \(-66.6\ \text{J}\), and choosing the loading platform as \(h=0\) would not change that result.

Using the ground as the reference,

\[
\Delta h=0.80-2.5=-1.70\ \text{m}
\]

so

\[
\begin{aligned}
\Delta U&=mg\Delta h\\
&=(4.0)(9.8)(-1.70)\\
&=-66.64\ \text{J}
\end{aligned}
\]

Therefore,

\[
\boxed{\Delta U\approx -67\ \text{J}}
\]

The negative sign means the gravitational potential energy decreases as the crate moves downward.

If the platform is chosen as \(h=0\), the crate begins at \(0\ \text{m}\) and finishes \(1.70\ \text{m}\) below it:

\[
h_i=0,\qquad h_f=-1.70\ \text{m}
\]

Therefore,

\[
\Delta h=-1.70-0=-1.70\ \text{m}
\]

which gives exactly the same \(\Delta U\).

The zero level affects the values assigned to \(U_i\) and \(U_f\), but not their difference.

Question 3

Two identical \(2.0\ \text{kg}\) objects start on the ground and finish on the same balcony \(6.0\ \text{m}\) above the ground. Object A is lifted vertically. Object B is pushed up a \(20\ \text{m}\) ramp.

A student claims that object B gains more gravitational potential energy because it travels further. Determine whether the claim is correct and calculate the gravitational potential energy gained by each object.

Solution 3

The claim is incorrect. Both objects gain \(117.6\ \text{J}\) of gravitational potential energy.

For each object,

\[
\Delta h=6.0\ \text{m}
\]

so

\[
\begin{aligned}
\Delta U&=mg\Delta h\\
&=(2.0)(9.8)(6.0)\\
&=117.6\ \text{J}
\end{aligned}
\]

Therefore,

\[
\boxed{\Delta U_A=\Delta U_B\approx 1.2\times10^2\ \text{J}}
\]

Gravitational potential energy depends on the change in vertical position, not the distance travelled along the path.

The \(20\ \text{m}\) ramp length may matter when calculating work done by other forces. For example, friction acting along the ramp could make the required input energy larger. But it does not change the gravitational potential energy gained between the same two heights.

Question 4

A \(1.5\ \text{kg}\) sensor is initially \(4.0\ \text{m}\) above a laboratory floor. A physicist chooses the sensor’s initial position as the zero of gravitational potential energy.

The sensor is lowered to a position \(1.0\ \text{m}\) above the floor.

Find:

a. the final height coordinate relative to the chosen reference level
b. the final gravitational potential energy relative to that reference
c. the change in gravitational potential energy

Explain why a negative value in part b is physically acceptable.

Solution 4

The final coordinate is \(-3.0\ \text{m}\), the final gravitational potential energy is \(-44.1\ \text{J}\), and the change in gravitational potential energy is also \(-44.1\ \text{J}\).

The chosen zero level is \(4.0\ \text{m}\) above the laboratory floor. The sensor finishes at \(1.0\ \text{m}\) above the floor, which is \(3.0\ \text{m}\) below the chosen reference.

Therefore,

\[
\boxed{h_f=-3.0\ \text{m}}
\]

Because the initial position was chosen as \(h_i=0\),

\[
U_i=0
\]

The final potential energy is

\[
\begin{aligned}
U_f&=mgh_f\\
&=(1.5)(9.8)(-3.0)\\
&=-44.1\ \text{J}
\end{aligned}
\]

so

\[
\boxed{U_f=-44.1\ \text{J}}
\]

The change is

\[
\begin{aligned}
\Delta U&=U_f-U_i\\
&=-44.1-0\\
&=-44.1\ \text{J}
\end{aligned}
\]

Therefore,

\[
\boxed{\Delta U=-44.1\ \text{J}}
\]

The negative value of \(U_f\) is acceptable because the zero of gravitational potential energy was chosen arbitrarily at the sensor’s starting height. Negative \(U\) simply means the sensor is below that reference.

The deeper misconception would be to interpret \(U=0\) as meaning “no gravitational interaction”. Gravity still acts at the reference level. Zero is just the value assigned for comparison.

Question 5

A \(0.60\ \text{kg}\) ball moves from point P to point Q. Relative to the floor, P is \(2.0\ \text{m}\) high and Q is \(7.0\ \text{m}\) high.

Student A chooses the floor as \(U=0\) and calculates

\[
U_P=11.76\ \text{J}, \qquad U_Q=41.16\ \text{J}
\]

Student B chooses point Q as \(U=0\) and says that the ball’s gravitational potential energy at P is \(-29.4\ \text{J}\).

The students argue that their results cannot both describe the same physical situation.

Evaluate the disagreement.

Solution 5

Both students’ calculations are consistent with the same physical situation because they have chosen different reference levels. Both predict the same increase in gravitational potential energy, \(+29.4\ \text{J}\).

For Student A,

\[
\begin{aligned}
\Delta U&=U_Q-U_P\\
&=41.16-11.76\\
&=29.40\ \text{J}
\end{aligned}
\]

For Student B, point Q is defined as \(U_Q=0\). Since P is \(5.0\ \text{m}\) below Q,

\[
h_P=-5.0\ \text{m}
\]

and therefore

\[
\begin{aligned}
U_P&=mgh_P\\
&=(0.60)(9.8)(-5.0)\\
&=-29.4\ \text{J}
\end{aligned}
\]

The change from P to Q is then

\[
\begin{aligned}
\Delta U&=U_Q-U_P\\
&=0-(-29.4)\\
&=+29.4\ \text{J}
\end{aligned}
\]

Thus,

\[
\boxed{\Delta U=+29.4\ \text{J}}
\]

for both descriptions.

The important test is not whether two observers assign the same numerical value to \(U\). They do not have to. The physically meaningful comparison is whether they predict the same change in energy between the same two positions.

13Where this idea leads next

Once \(\Delta U=mg\Delta h\) is clear, gravitational energy problems become much easier to organise.

When an object falls, its gravitational potential energy decreases. If air resistance and other dissipative effects are negligible, that decrease can appear as an increase in kinetic energy. That gives the next major tool:

\[
\Delta K+\Delta U=0
\]

for a system in which mechanical energy is conserved.

The reference level still does not matter. Changing the zero of \(U\) shifts the potential energy values, but the energy changes, the predicted speeds, and the physics remain the same.

Later, when the distances become large enough that \(g\) can no longer be treated as constant, the same reference-level idea survives, but the simple near-Earth expression \(mg\Delta h\) is replaced by the full gravitational potential energy model.