Potential Energy Calculator

potential energy = mass × gravitational acceleration × height lifted

m

The mass being lifted, in kilograms, with grams, tonnes and pounds in the same field. It scales the energy in a straight line: twice the mass, twice the joules, no matter how high you go. It does not appear in the falling speed at all — that is the one place on this page where mass drops out, and it is worth watching for

h

How far the object is raised, in metres, feet, centimetres or inches. It is a difference and not an altitude: carrying a box from the first floor to the third makes h the two-storey gap of about six metres, not the height above sea level and not the distance to the centre of the Earth. Zero is allowed and is a real reading — no lift, no stored energy

g

The gravitational acceleration where the lifting happens, in m/s², ft/s² or as a multiple of standard gravity. This is the field that makes the page worth opening: it defaults to 9.80665, and the table beside the calculator lists six bodies so the value can be copied straight in. The Moon's 1.62 is a sixth of Earth's, so the same lift stores a sixth of the energy; the Sun's 274 is twenty-eight times Earth's, and a 70 kg person would weigh 19.2 kN there

U

The stored energy, in joules as the main row, with kilojoules and foot-pounds underneath for readers who work in either. It is the work gravity will give back when the object comes down, and the work you put in to raise it — which is why it is also called gravitational potential energy rather than just potential energy. On Earth a 70 kg person raised 2 m stores 1372.93 J, about a third of a food calorie

√(2gh)

The speed the object would reach if it were dropped from that height — its impact velocity — in m/s. It is the same energy seen from the other end: all of the stored U becomes kinetic energy on the way down, and because both sides of ½mv² = mgh carry an m, the mass cancels and the answer depends only on height and gravity. Raising an object twice as high makes it land 1.41 times as fast, not twice as fast

Use this page when you need to know what a lift costs or what a drop delivers: sizing a hoist or a motor for a crane, working out the energy in a pumped-storage reservoir or a raised weight clock, comparing what a battery has to supply to raise a load, or answering the classroom question of where the energy goes when a ball is held above the ground. The gravity field is what separates this page from a plain mgh multiplication — set it to 1.62 and the same lift costs a sixth as much, which is the whole reason lunar and Mars mission planning cares about this number. The second reason to open it is the falling speed: if you have a height and you want to know how hard something lands, the mass you may not know does not matter.

1. The defaults: 70 kg raised 2 m on Earth

   1. Mass 70 kg, height 2 m, gravity 9.80665 m/s²
   2. Energy: U = m × g × h = 70 × 9.80665 × 2 = 1372.931 J, which prints as 1372.93 J
   3. In kilojoules: 1372.931 ÷ 1000 = 1.372931, which prints as 1.3729 kJ
   4. In foot-pounds: 1372.931 ÷ 1.3558179483 = 1012.6219, which prints as 1012.62 ft·lb
   5. Falling speed: √(2 × 9.80665 × 2) = √39.2266 = 6.2631, which prints as 6.26 m/s
   6. Sanity check on the last row: ½ × 70 × 6.2631² = 1372.9 J — the falling speed is the same energy, not a separate fact

   The energy is about a third of a food calorie, which is the useful scale to have in mind: lifting a person up one storey costs roughly a third of a Calorie of mechanical work, and your body spends several times that because muscles are not 100 percent efficient. The last step is the reason the falling speed belongs on this page at all. It is not an extra formula bolted on — it is the same 1372.93 J written as motion instead of as height, and the check confirms it. Notice also what the falling speed does not contain: no mass. Drop a 70 kg person and a 1 kg bag from 2 m and both hit at 6.26 m/s, which is why the height in a fall tells you the landing speed and the weight only tells you the damage.

2. A bottle of water raised to a desk: 1 kg, 1 m

   1. Mass 1 kg, height 1 m, gravity 9.80665 m/s²
   2. Energy: U = 1 × 9.80665 × 1 = 9.80665 J, which prints as 9.81 J
   3. In kilojoules: 0.00980665, which prints as 0.0098 kJ — four decimals, not two
   4. In foot-pounds: 9.80665 ÷ 1.3558179483 = 7.2330, which prints as 7.23 ft·lb
   5. Falling speed: √(2 × 9.80665 × 1) = √19.6133 = 4.4287, which prints as 4.43 m/s

   This is the row that explains why the kilojoule result keeps four decimals instead of two. At this scale the two-decimal row would read 0.01 kJ, which is both wrong in the second digit and inconsistent with the joule row printed next to it — 9.81 J is 0.00981 kJ, and the panel says so. The falling speed is worth a second look too: 4.43 m/s is about 16 km/h, which is a brisk run, and it is what a bottle reaches falling off a desk-height shelf. Small heights give small energies, and the square root means they give less small speeds than you might expect.

3. The same lift on the Moon: 70 kg, 2 m, g = 1.62

   1. Mass 70 kg, height 2 m, gravity 1.62 m/s² — the Moon
   2. Energy: U = 70 × 1.62 × 2 = 226.8 J, exactly a sixth of the 1372.93 J it costs on Earth
   3. In kilojoules: 0.2268, which prints as 0.2268 kJ
   4. In foot-pounds: 226.8 ÷ 1.3558179483 = 167.2791, which prints as 167.28 ft·lb
   5. Falling speed: √(2 × 1.62 × 2) = √6.48 = 2.5456, which prints as 2.55 m/s

   Every row shrinks by the ratio of the two gravities, 1.62 ÷ 9.80665 = 0.1652, and that ratio is the whole point of making gravity an input instead of writing 9.81 into the formula. The falling speed shrinks too, but by the square root of that ratio rather than by the ratio itself — 6.2631 × √0.1652 = 2.5456, which prints as 2.55 — and that difference between a linear and a square-root response is worth noticing. An astronaut in the Apollo footage hops rather than walks for this reason: the same leg push that lifts them 20 cm on Earth lifts them 1.2 m here, and landing from that height is slower and softer.

The height is a difference between two levels, so the answer is the energy of that particular lift, not the total energy an object has by position: the reference level is wherever you say it is. Gravity is constant over the whole lift, exact over a building and wrong over a distance comparable to a planet's radius. Nothing here covers sideways motion: the object starts and ends at rest. Air resistance is ignored in the falling speed, along with the Earth's rotation and the buoyancy of the air.

What is the potential energy formula?

U = m × g × h: mass times gravitational acceleration times height lifted. A 70 kg person raised 2 m on Earth stores 70 × 9.80665 × 2 = 1372.93 J. The formula is one multiplication, which is why the interesting part of this page is not the arithmetic but the two things it lets you vary — the gravity, which turns the same lift into a sixth of the energy on the Moon, and the height, which is a difference between two levels rather than an altitude.

Why does the impact velocity not depend on the mass?

Because the mass appears on both sides of the energy balance and cancels. Falling converts stored energy into kinetic energy, so m × g × h = ½ × m × v², and dividing both sides by m leaves g × h = ½v², which rearranges to v = √(2gh). A 70 kg person and a 1 kg bag dropped from 2 m both land at 6.26 m/s. Heavier objects carry more energy — that is the 1372.93 J — but they also need more energy to be sped up, and the two effects cancel exactly.

What does raising an object twice as high do?

It doubles the stored energy but multiplies the falling speed by only 1.41, the square root of two. Going from 2 m to 4 m takes a 70 kg lift from 1372.93 J to 2745.86 J, while the landing speed goes from 6.26 m/s to 8.86 m/s rather than to 12.5. This is worth internalising because it is the same relationship that makes braking distances grow with the square of speed: the energy tracks the height, and the speed only tracks its square root.

What gravity value should I use for the Moon or Mars?

The table beside the calculator lists both: the Moon is 1.62 m/s² and Mars is 3.71, against Earth's defined 9.80665. Those are surface averages and vary by about a percent across each body, which is far below the differences between them, so they are the right values for sizing missions. Jupiter is listed as 24.79 m/s², but note that its surface is a 1-bar pressure level rather than solid ground, and the Sun's 274 is its photosphere — a 70 kg person would weigh 19.2 kN there.

Is potential energy the same as kinetic energy?

They are the same quantity in two forms, and this page's second half is the exchange rate. Energy stored by height is released as energy of motion on the way down, so the 1372.93 J of a 70 kg lift at 2 m becomes 1372.93 J of kinetic energy at the moment before landing, which is exactly the 6.26 m/s the page reports. The kinetic energy calculator starts from the other end and gives the height needed for a given speed, so the two pages are inverses of each other rather than duplicates.

• Gravitational Potential Energy (College Physics 2e, §7.3) — U = mgh, the choice of reference level, and the conversion between stored energy and falling speed — OpenStax
• Gravitational energy — the mgh form, its limits, and why the mass cancels out of the impact speed — Wikipedia
• NIST Guide for the Use of the International System of Units (SI), Appendix B.8 — Factors for Units Listed Alphabetically: the foot pound-force row (1 ft·lbf = 1.355 818 J) that this page's third row is converted with — NIST