91影视

a) Weight of the object on the surface of earth is$W=100\text{鈥凬}$

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

c) Gravitational constant,$G=6.67脳{10}^{鈭�11}\text{N}鈰�{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Newton鈥檚 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鈥檚 weight is the same as that on the moon鈥檚 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

$\begin{array}{c}{g}_m=\left(\frac{1}{6}\right)g_e\\ =\left(\frac{1}{6}\right)\left(9.8{\text{鈥夆赌夆赌尘/蝉}}^{\text{2}}\right)\text{(Given)}\\ =1.63{\text{鈥刴/蝉}}^{\text{2}}\end{array}$

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

$\begin{array}{c}{W}_e=m{g}_e\\ 100\text{鈥夆赌塏}=m\left(9.8{\text{鈥夆赌尘/蝉}}^{\text{2}}\right)\\ m=\frac{100\text{鈥夆赌塏}}{9.8{\text{鈥夆赌夆赌尘/蝉}}^{\text{2}}}脳\left(\frac{\text{1鈥夆赌塳驳}鈰�{\text{m/s}}^{\text{2}}}{1\text{鈥夆赌塏}}\right)\\ =10.2\text{鈥刱驳}\end{array}$

So the weight of the object on the moon is,

$\begin{array}{c}{W}_m=10.2\text{鈥嬧�刱驳}脳1.63{\text{鈥刴/蝉}}^{\text{2}}\\ =16.7\text{鈥凬}\\ 鈮�17\text{鈥凬}\end{array}$

Hence, the weight of the object is approximately $17\text{鈥凬}$

Here the weight on the moon is equal to Earth鈥檚 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,

$\begin{array}{c}r=\sqrt{\frac{G{M}_e}{g_m}}\\ =\sqrt{\frac{6.67脳{10}^{鈭�11}\text{鈥凬}鈰�{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}脳5.97脳{10}^{24}\text{鈥刱驳}}{1.63{\text{鈥刴/蝉}}^{\text{2}}}}\\ =\sqrt{24.4293脳{10}^{13}{\text{鈥刴}}^{\text{2}}}\\ =1.56脳{10}^7\text{鈥刴}\end{array}$

So in terms of radius of earth we get,

$\begin{array}{c}r=\frac{1.56脳{10}^7\text{鈥刴}}{r_e}\\ =\frac{1.56脳{10}^7\text{鈥刴}}{6.4脳{10}^6\text{鈥刴}}\\ =2.4\text{radii}\end{array}$

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