490.5
J
0.4905
kJ
9.9
m/s
490.5
J
0.4905
kJ
9.9
m/s
Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. When you lift a book off the floor, you do work against gravity, and that energy is stored as potential energy ready to be converted back into kinetic energy if the book falls. The formula $$PE = mgh$$ quantifies this stored energy for objects near Earth's surface, where the gravitational field is approximately uniform.
This Potential Energy Calculator determines the gravitational PE for any mass at a given height, and also computes the impact velocity the object would reach if dropped from that height (assuming no air resistance). The relationship $$v = \sqrt{2gh}$$ comes directly from energy conservation: all potential energy converts to kinetic energy during free fall. Engineers use these calculations for structural load analysis, dam design, and safety assessments. Athletes and coaches apply them to understand jumping mechanics and fall injuries.
The gravitational potential energy near Earth's surface is:
$$PE = mgh$$
where $$m$$ is mass (kg), $$g$$ is gravitational acceleration (default 9.81 m/s²), and $$h$$ is height above the reference point (m). The default gravity value applies at sea level on Earth; you can adjust it for other planets or altitudes. If the object is released from rest, conservation of energy gives $$mgh = \frac{1}{2}mv^2$$, yielding the impact velocity $$v = \sqrt{2gh}$$. Note that this velocity is mass-independent — a feather and a bowling ball hit the ground at the same speed in a vacuum.
A PE of 100 J means the object can do 100 joules of work as it falls to the reference level. For context, a 70 kg person standing on a 3 m ledge has about 2,060 J of PE — equivalent to the kinetic energy of a car bumper impact at low speed. The impact velocity helps assess fall hazards: a fall from 5 m yields 9.9 m/s (about 35 km/h), which is potentially fatal for humans.
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A 2 kg textbook at 1.5 m height has PE = 2 × 9.81 × 1.5 = 29.43 J. If dropped, it hits at 5.42 m/s.
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1 cubic meter of water (1000 kg) at 20 m height stores 196.2 kJ of potential energy — the basis of hydroelectric power.
The reference point (where h = 0) is arbitrary and chosen for convenience. Common choices are the ground, the floor, or the lowest point in the problem. Only differences in potential energy are physically meaningful, so the reference point cancels out in calculations.
Yes, but you must use the local gravitational acceleration. For the Moon, g ≈ 1.62 m/s². For Mars, g ≈ 3.72 m/s². For Jupiter, g ≈ 24.79 m/s². The advanced input lets you adjust gravity for any celestial body.
This formula assumes a uniform gravitational field, which is valid near Earth's surface. For very large heights (comparable to Earth's radius, 6,371 km), you must use the general formula $$PE = -\frac{GMm}{r}$$ instead, where gravity varies with distance.
With $$PE = mgh$$, it can be negative if the object is below the reference point (negative h). In the universal form $$PE = -GMm/r$$, potential energy is always negative and approaches zero at infinity. The sign convention depends on the reference point chosen.
In a conservative system (no friction/air resistance), total mechanical energy is conserved: $$PE + KE = \text{constant}$$. As an object falls, PE decreases and KE increases by the same amount. At ground level, all PE has converted to KE: $$mgh = \frac{1}{2}mv^2$$.
A 70 kg person at 10 m: $$PE = 70 \times 9.81 \times 10 = 6867$$ J ≈ 6.87 kJ. Upon impact with the water, they would be traveling at $$v = \sqrt{2(9.81)(10)} ≈ 14$$ m/s (about 50 km/h) if air resistance is neglected.
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