How can I calculate gravitational potential-energy change near a uniform field?
Quick answer
Near Earth’s surface, calculate the change in gravitational potential energy with ΔUg = mgΔh. Use mass in kilograms, g as a positive value in m/s², and signed height change Δh in metres. An upward change is positive and a downward change is negative. This assumes g is effectively constant over the height change; your chosen zero-height reference does not affect the change.
The question, the details, the next step.
What formula should I use for a height change near Earth’s surface, and what exactly counts as Δh?
For the near-surface, uniform-field model, use ΔUg = mgΔh. Here, m is the object’s mass, g is the positive magnitude of gravitational-field strength, and Δh = h_final − h_initial is the signed vertical change. This relation applies near Earth’s surface when g is treated as constant. Only differences in gravitational potential energy are physically significant, so you can choose a reference height and assign it zero potential energy [S1].
For a rise of 2 m, Δh = +2 m and the change is positive. For a fall of 2 m, Δh = −2 m and the change is negative. Use the difference in vertical height between the starting and ending positions.
Could you work through a calculation and show the units?
Suppose, as an example, a 2.0 kg object is raised 3.0 m. Take g = 9.8 m/s² for this calculation and set the starting height as the reference. Then Δh = +3.0 m:
- Substitute: ΔUg = (2.0 kg)(9.8 m/s²)(+3.0 m).
- Multiply: ΔUg = +58.8 kg·m²/s².
- Since 1 kg·m²/s² = 1 joule, ΔUg = +58.8 J, or about +59 J to two significant figures.
The positive sign means the object’s gravitational potential energy increased. This agrees with the 2 kg, 3 m example, which gives approximately 59 J [S1]. The result’s precision depends on the values used and how they are rounded.
What if the object moves downward, or I choose a different zero height?
For a descent, keep g positive and make Δh negative. For example, if the same 2.0 kg object moves down 3.0 m, then ΔUg = (2.0)(9.8)(−3.0) J = −58.8 J. The negative result indicates a decrease in gravitational potential energy; the downward direction is represented by Δh, not by making g negative [S1].
Changing the zero-height reference changes the potential-energy values assigned at individual positions, but not their difference, provided you use the same reference for both. For example, calling the floor zero or calling a tabletop zero gives the same ΔUg for the same initial and final positions [S1].
How can I check my own answer, and when should I be cautious about this formula?
Check that you subtracted the initial height from the final height and kept the sign. Confirm that the units reduce to joules: kg × m/s² × m = kg·m²/s². Then check the direction: a rise should give a positive change, and a fall a negative one.
This formula uses the near-surface idealization that g is constant over the height change [S1]. The formula alone does not establish that this approximation is suitable for every distance or gravitational setting. If a problem specifies that g varies, or gives another model, follow the method and field information provided for that situation rather than assuming this calculation applies.
A practical checklist
Write down the initial and final vertical heights, using the same reference.
Calculate Δh = h_final − h_initial and retain its sign.
Record mass in kilograms and g as a positive value in m/s².
Substitute into ΔUg = mgΔh and show the units before converting to joules.
Check that the sign agrees with whether the object rose or fell.
A few more useful answers
Does the chosen zero height affect ΔUg?
No. It changes the potential-energy values assigned to individual positions, but not the difference between them when the same reference is used for both [S1].
Can gravitational potential energy itself be negative?
Its assigned value depends on the chosen zero reference. For this calculation, focus on ΔUg, the difference between final and initial values; changing the reference does not change that difference [S1].
Why is g not negative when the object moves downward?
In this formula, g is the positive magnitude of the field strength, while the direction of the height change is carried by the sign of Δh. A downward change therefore produces a negative ΔUg [S1].
Sources
References used for this guide.
- [S1] OpenStax College Physics 2e, Chapter 7 Section Summary: Gravitational Potential Energy openstax.org · Checked