Weight of an object on the Moon

An astronaut who weighs 588 N standing on Earth would weigh only about 98 N standing on the Moon — same body, same mass, yet a very different pull of gravity underfoot. This puzzle sits inside the CBSE Class 9 chapter Gravitation, and it is one of the cleanest tests of whether a student truly understands the difference between mass and weight. Mass is how much matter a body contains, and it never changes just because the body travels somewhere else. Weight is the force of gravity pulling on that mass, $W = mg$ , and $g$ is different on every world because it depends on that world's own mass and radius. The Moon is smaller and far less massive than Earth, so its gravitational pull is weaker — and the numbers work out to almost exactly one-sixth. This article walks through why that one-sixth ratio appears, and how to use it confidently in numericals.

The core insight

Because mass never changes with location but weight equals $mg$ , and the Moon's $g \approx 1.63\ \text{m/s}^2$ is about one-sixth of Earth's $g \approx 9.8\ \text{m/s}^2$ , any object's weight on the Moon is approximately one-sixth of its weight on Earth.

The core idea

Acceleration due to gravity at the surface of any body is given by $g = \dfrac{GM}{R^2}$ , where $G$ is the universal gravitational constant, $M$ is the mass of the attracting body, and $R$ is its radius. Because $M$ and $R$ are different for every planet or moon, $g$ is different everywhere too.

g_{\text{Moon}} \approx 1.63\ \text{m/s}^2 \qquad g_{\text{Earth}} \approx 9.8\ \text{m/s}^2

where: $g_{\text{Moon}}$ and $g_{\text{Earth}}$ are the accelerations due to gravity at the surfaces of the Moon and Earth respectively, both measured in metres per second squared. Dividing one by the other gives the famous ratio used throughout this chapter:

\frac{g_{\text{Moon}}}{g_{\text{Earth}}} = \frac{1.63}{9.8} \approx \frac{1}{6}

Since weight is defined as $W = mg$ , and an object's mass $m$ stays exactly the same wherever it goes, its weight scales in direct proportion to whatever value of $g$ it happens to be sitting in:

W_{\text{Moon}} = m \times g_{\text{Moon}} \approx \frac{1}{6}\,m \times g_{\text{Earth}} = \frac{1}{6}\,W_{\text{Earth}}

where: $W_{\text{Moon}}$ and $W_{\text{Earth}}$ are the weights (in newtons) of the same object on the Moon and on Earth, and $m$ is its mass (in kilograms), unchanged in both places. So an object's weight on the Moon is roughly one-sixth of its weight on Earth, even though its mass is identical in both places. This is exactly why astronauts in bulky spacesuits can leap high off the lunar surface and carry equipment that would be exhausting on Earth — their muscles fight a much smaller gravitational force, even though their own inertia (which depends on mass, not weight) is unchanged. It helps to see why the Moon's gravity ends up weaker rather than stronger. The Moon's mass is only about $1/81$ of Earth's mass, while its radius is about $0.27$ times Earth's radius. A smaller radius alone would tend to increase $g$ , since $g \propto 1/R^2$ rewards being closer to the centre of mass. But the collapse in mass is so much larger a factor that it dominates completely, and the net result is a much smaller $g$ on the Moon. The lesson is that $g$ is never simply "smaller for a smaller world" — it depends on the precise combination of mass and radius, and for the Moon the mass deficit wins decisively.

Figure

Bar chart comparing the weight of a 72 kilogram astronaut on Earth and on the Moon showing the Earth bar about six times taller than the Moon bar

Derivation AWhy weight on the Moon is about one-sixth of weight on Earth

The goal of this derivation is to show, starting only from the definitions $g = GM/R^2$ and $W = mg$ together with the Moon's actual mass and radius relative to Earth's, exactly why an object's weight on the lunar surface comes out to close to one-sixth of its weight on Earth rather than some other fraction.

Step 1. Write $g$ for each body using $g = GM/R^2$ : $g_{\text{Earth}} = \dfrac{GM_{\text{Earth}}}{R_{\text{Earth}}^2}$ and $g_{\text{Moon}} = \dfrac{GM_{\text{Moon}}}{R_{\text{Moon}}^2}$ . The constant $G$ is the same everywhere in the universe, so it will cancel when we take a ratio.

Step 2. Form the ratio $g_{\text{Moon}}/g_{\text{Earth}}$ , which cancels $G$ : $\dfrac{g_{\text{Moon}}}{g_{\text{Earth}}} = \dfrac{M_{\text{Moon}}}{M_{\text{Earth}}} \times \left(\dfrac{R_{\text{Earth}}}{R_{\text{Moon}}}\right)^2$ .

Step 3. Substitute the known values $M_{\text{Moon}}/M_{\text{Earth}} \approx 1/81$ and $R_{\text{Moon}}/R_{\text{Earth}} \approx 0.27$ , so $R_{\text{Earth}}/R_{\text{Moon}} \approx 3.7$ : $\dfrac{g_{\text{Moon}}}{g_{\text{Earth}}} \approx \dfrac{1}{81} \times (3.7)^2 \approx \dfrac{1}{81} \times 13.7 \approx 0.169$ , which is very close to the accepted ratio $1.63/9.8 \approx 0.166$ .

Step 4. Multiply both sides by $W_{\text{Earth}} = m g_{\text{Earth}}$ to convert the $g$ ratio into a weight ratio: since $W_{\text{Moon}} = m g_{\text{Moon}}$ and $W_{\text{Earth}} = m g_{\text{Earth}}$ share the same $m$ , dividing gives $W_{\text{Moon}}/W_{\text{Earth}} = g_{\text{Moon}}/g_{\text{Earth}} \approx 1/6$ , so $W_{\text{Moon}} \approx W_{\text{Earth}}/6$ — the mass cancels out completely, confirming the ratio depends only on the two worlds' gravity, not on how heavy the object itself is.

Worked example

An astronaut has a mass of 72 kg. Find her weight on (a) the Earth, and (b) the Moon, taking $g_{\text{Earth}} = 9.8\ \text{m/s}^2$ and $g_{\text{Moon}} = 1.63\ \text{m/s}^2$ .

Step 1. Weight on Earth: $W_{\text{Earth}} = mg_{\text{Earth}} = 72 \times 9.8 = 705.6\ \text{N}$ .

Step 2. Weight on the Moon: $W_{\text{Moon}} = mg_{\text{Moon}} = 72 \times 1.63 = 117.36\ \text{N}$ .

Step 3. Cross-check with the one-sixth rule: $705.6 / 6 = 117.6\ \text{N}$ , which closely matches the direct calculation (the tiny difference exists because $1.63/9.8$ is only approximately, not exactly, $1/6$ ). Her mass, however, remains exactly 72 kg on both worlds.

Final answer: the astronaut weighs 705.6 N on Earth and 117.36 N on the Moon, while her mass stays 72 kg in both places.

Competitive stretch

A rock weighs 480 N on Earth's surface. Find its mass, and its weight on the Moon. First find mass from $m = W_{\text{Earth}}/g_{\text{Earth}} = 480/9.8 = 48.98\ \text{kg} \approx 49\ \text{kg}$ — this mass does not change on the Moon. Then $W_{\text{Moon}} = m \times g_{\text{Moon}} = 48.98 \times 1.63 \approx 79.8\ \text{N}$ . Using the quick one-sixth rule directly on the Earth weight gives $480/6 = 80\ \text{N}$ , which agrees closely. The rock's weight on the Moon is about 80 N.

Common traps

Assuming mass changes on the Moon. Mass never changes with location — only weight does, because weight depends on the local value of $g$ , while mass is a fixed property of the object.
Using $g_{\text{Earth}}$ by mistake in a Moon-weight problem, or forgetting to switch to $g_{\text{Moon}} = 1.63\ \text{m/s}^2$ in the formula $W = mg$ .
Treating "one-sixth" as an exact ratio. The precise ratio $g_{\text{Moon}}/g_{\text{Earth}} \approx 1.63/9.8 \approx 0.166$ is close to but not exactly $1/6 = 0.1667$ — fine for quick estimation, but a numerical with given $g$ values should use those exact values.
Confusing "weaker gravity" with "no gravity" on the Moon. The Moon does have its own gravitational pull, which is why astronauts stay on its surface instead of floating away; it is simply about six times weaker than Earth's because the Moon has far less mass.

Ask the tutor: Want to work out your own approximate weight on the Moon, or see how the same $g = GM/R^2$ logic plays out on Mars or Jupiter?

If I weigh less on the Moon, does that mean I am lighter, as in less matter?

No. "Lighter" in everyday speech is misleading here. The amount of matter in your body, your mass, is exactly the same on the Moon as on Earth — it does not depend on gravity at all. What changes is weight, the force with which gravity pulls on that mass, because the Moon's gravitational field is weaker. You would still need the same effort to push a stalled car sideways on the Moon (that resists mass, via inertia) even though lifting it straight up would feel far easier (that resists weight).

Why is the ratio close to one-sixth and not some other fraction like one-third or one-tenth?

It comes directly from the Moon's actual mass and radius compared to Earth's, through $g = GM/R^2$ . The Moon's mass is about $1/81$ of Earth's, which on its own would make $g$ eighty-one times smaller. But the Moon's radius is also smaller, about $0.27$ times Earth's, and a smaller radius pulls objects closer to the centre, increasing $g$ by a factor of about $(1/0.27)^2 \approx 13.7$ . Multiplying $1/81$ by $13.7$ gives approximately $1/6$ — a coincidence of the Moon's particular size and mass, not a universal law that every smaller moon or planet obeys.

Can I use the one-sixth rule for every Moon-related numerical, or do I need the derivation?

For quick estimates or to sanity-check an answer, the one-sixth rule is perfectly fine and is exactly what most CBSE numericals expect. But when a question explicitly gives you $g_{\text{Moon}} = 1.63\ \text{m/s}^2$ and $g_{\text{Earth}} = 9.8\ \text{m/s}^2$ , you should calculate directly with $W = mg$ using those values, since $1.63/9.8$ is only approximately $1/6$ . Use the one-sixth rule as a mental check on your final answer, not as a replacement for the exact calculation when precise values are given.

Weight of an object on the Moon

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads