(a) What will an object weigh on the Moon’s surface if it weighs $100\text{N}$on Earth’s surface? (b) How many Earth radii must this same object be from the centre of Earth if it is to weigh the same as it does on the Moon?

a) Weight of the object on the surface of earth is$W=100\text{\hspace{0.33em}N}$

b) Acceleration due to gravity on the Earth,$g_e=9.8{\text{m/s}}^{\text{2}}$

c) Gravitational constant,$G=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Newton’s law of gravitation states that any particle in the universe attracts any other particle with a gravitational force whose magnitude is

$F=G\frac{m_1{m}_2}{r^2}$

Here,$m_1$androle="math" localid="1655784760791" $m_2$are masses of the particles and $r$is their separation and $G$is the gravitational constant.

We use the concept of gravitational force to find the weight of the object on the moon. Also, we find the distance which is multiple radii of earth where the object’s weight is the same as that on the moon’s surface.

Formulae:

Weight of an object, $W=mg$ ...(i)

Gravitational Force, $F=\frac{GMm}{r^2}$ ...(ii)

We know the gravity of the moon is

$\begin{array}{c}{g}_m=\left(\frac{1}{6}\right)g_e\\ =\left(\frac{1}{6}\right)\left(9.8{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m/s}}^{\text{2}}\right)\text{(Given)}\\ =1.63{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

We can find the mass of the object from the weight on the earth using equation (i),

$\begin{array}{c}{W}_e=m{g}_e\\ 100\text{\hspace{0.17em}\hspace{0.17em}N}=m\left(9.8{\text{\hspace{0.17em}\hspace{0.17em}m/s}}^{\text{2}}\right)\\ m=\frac{100\text{\hspace{0.17em}\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m/s}}^{\text{2}}}\times \left(\frac{\text{1\hspace{0.17em}\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}}{1\text{\hspace{0.17em}\hspace{0.17em}N}}\right)\\ =10.2\text{\hspace{0.33em}kg}\end{array}$

So the weight of the object on the moon is,

$\begin{array}{c}{W}_m=10.2\text{\hspace{0.33em}kg}\times 1.63{\text{\hspace{0.33em}m/s}}^{\text{2}}\\ =16.7\text{\hspace{0.33em}N}\\ \approx 17\text{\hspace{0.33em}N}\end{array}$

Hence, the weight of the object is approximately $17\text{\hspace{0.33em}N}$

Here the weight on the moon is equal to Earth’s gravitational force

$\begin{array}{c}{W}_m=\frac{G{M}_{e}m}{r^2}\text{(usingequation(ii))}\\ m{g}_m=\frac{G{M}_{e}m}{r^2}\text{(usingequation(i))}\\ g_m=\frac{G{M}_e}{r^2}\end{array}$

Rearranging it for$r_e$we get,

$\begin{array}{c}r=\sqrt{\frac{G{M}_e}{g_m}}\\ =\sqrt{\frac{6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 5.97\times {10}^{24}\text{\hspace{0.33em}kg}}{1.63{\text{\hspace{0.33em}m/s}}^{\text{2}}}}\\ =\sqrt{24.4293\times {10}^{13}{\text{\hspace{0.33em}m}}^{\text{2}}}\\ =1.56\times {10}^7\text{\hspace{0.33em}m}\end{array}$

So in terms of radius of earth we get,

$\begin{array}{c}r=\frac{1.56\times {10}^7\text{\hspace{0.33em}m}}{r_e}\\ =\frac{1.56\times {10}^7\text{\hspace{0.33em}m}}{6.4\times {10}^6\text{\hspace{0.33em}m}}\\ =2.4\text{radii}\end{array}$

The object should have at 2.4radii from the centre of earth