Question

the acceleration due to gravity at moon surface is 1.67m/s.if radius of moon is 1.74.10^6m.calculate the mass of moon.

Answer 1

Hi mate.

Thanks Thanks for asking this question.

Here is your answer,

Given :

$g \: = \: 1.67 \frac{m}{s} \\ \\ r = \: 1.74 \times {10}^{6} m$
${g}^{o} = universal \: gravitational \: \\ \: \: \: \: \: \: \: constant \: = 6.673 \times {10}^{ - 11}$

m = mass of moon

We know that,

$g = \frac{ {g}^{o}m }{ {r}^{2} }$

$1.67 = \frac{6.67 \times {10}^{ - 11} \times m }{ {(1.74 \times {10}^{6)} }^{2} }$

$m = \frac{1.67 \times {(1.74 \times }^{2} { {10}^{6}) }^{2} }{6.673 \times {10}^{ - 11} }$

$m = \frac{1.67 \times 3.0276 \times {10}^{12} }{6.673 \times {10}^{ - 11} }$

$m = \frac{5.056 \times {10}^{23} }{6.673}$

$m = 0.7576 \times {10}^{23}$

$m = 7.576 \times {10}^{22} \: kg$




The required mass of moon is
$7.576 \times {10}^{22} kg$



Thank you.




BE BRAINLY !!!!!!!!!!

Answer 2

Answer:

For the given data the mass of the moon will be equal to 7.58×10²²Kg.

Explanation:

Given that acceleration at moon surface due to gravity, $g=1.67ms^{-2}$

Radius of the moon, $R=1.74*10^{6}m$

We know acceleration at the moon, $g=\frac{GM}{R^{2}}$

⇒            $M=\frac{gR^{2}}{G}$                                    ..............(1)

Where $G=universal$ $gravitational$ $constant$

           $G=6.67*10^{-11}$

Put the value g, G and R in eq.(1),

We get      $M=\frac{(1.67)(1.74*10^{6})^{2}}{6.67*10^{-11}}$

⇒              $M=\frac{(1.67)(3.02*10^{12})}{6.673*10^{-11}}$

⇒             $M=\frac{(5.056*10^{23})}{6.673}}$

 ∴             $M=7.58*10^{22}kg$

Therefore the mass of moon is equal to 7.58×10²²Kg.