A body of mass m = 10 kg is taken from Earth to the Moon.…

A body of mass $m = 10\ \mathrm{kg}$ is taken from Earth to the Moon. Find: (a) the gravitational force on it on Earth and on the Moon; (b) the reading of a dynamometer at rest from which it hangs, on Earth and on the Moon; (c) its weight when it lies on the ground of each body; (d) its weight in a rocket launched from Earth, at the moment the rocket accelerates vertically upward with $a = 4g$; (e) its weight in a rocket launched from the Moon with the same acceleration. ($g_E = 9.81\ \mathrm{m/s^2}$, $g_M = 1.62\ \mathrm{m/s^2}$.)

Explanation of Method

Gravity, dynamometer reading and weight on Earth, the Moon and in a rocket.

Step 1: (a) Gravitational force

\[ F_E = m g_E = 98.1\ \mathrm{N}, \]

\[ F_M = m g_M = 16.2\ \mathrm{N}. \]

Step 2: (b) and (c) At rest

A body at rest has zero acceleration, so the dynamometer (or the ground) must balance gravity exactly. The weight, the force the body exerts on its support, is then equal to the gravitational force:

\[\text{Earth: } 98.1\ \mathrm{N}, \qquad \text{Moon: } 16.2\ \mathrm{N}\]

Step 3: (d) Rocket launched from Earth

The support pushes up with $N$, gravity pulls down with $mg_E$, and the body accelerates upward with $a$: $N - mg_E = ma$.

\[ N = m\,(g_E + a) = 10\,(9.81 + 39.24). \]

\[W = N = 490.5\ \mathrm{N} \ \ (= 5\,mg_E)\]

Step 4: (e) Rocket launched from the Moon

With the same acceleration $a = 4g = 39.24\ \mathrm{m/s^2}$: $N = m\,(g_M + a) = 10\,(1.62 + 39.24)$.

\[W \approx 408.6\ \mathrm{N}\]

$\text{gravity}$ $98.1$ $16.2$ $\text{dynamometer}$ $98.1$ $16.2$ $\text{on the ground}$ $98.1$ $16.2$ $\text{rocket, a = 4g}$ $490.5$ $408.6$ $\text{force (N)}$

Earth
Moon

Bars to scale. At rest the dynamometer and the ground both show the gravitational force. In a rocket accelerating upward the support must also provide $ma$, so the weight grows.

Result

\[\boxed{\,\text{Earth: } 98.1\ \mathrm{N}, \qquad \text{Moon: } 16.2\ \mathrm{N}\,}\]

\[\boxed{\,W = N = 490.5\ \mathrm{N} \ \ (= 5\,mg_E)\,}\]

\[\boxed{\,W \approx 408.6\ \mathrm{N}\,}\]

Remark

About the original solution. Parts (a)–(d) are correct. In (e) it used $a = 4g_M = 6.48\ \mathrm{m/s^2}$, giving 81 N; that answers "four times the Moon's gravity", not "the same acceleration" as the Earth launch. With the same acceleration, $4 \cdot 9.81\ \mathrm{m/s^2}$, the weight is 408.6 N. If the problem does mean $a = 4g_M$, then $W = 5\,mg_M = 81\ \mathrm{N}$.