Calculate the value of acceleration due to gravity on the surface of the moon.
in Physics
26 views
2 Votes
2 Votes
Calculate the value of acceleration due to gravity on the surface of the moon.

(Given Mass of the moon $=7.4 \times 10^{22} \mathrm{~kg} ;$ Radius of moon $\left.=1740 \mathrm{~km} ; G=6.7 \times 10^{-11} \mathrm{Nm}^{2}/\mathrm{kg}^{2}\right)$
in Physics
by
7.2k Points

1 Answer

2 Votes
2 Votes
 
Best Answer

Solution

The formula for calculating the acceleration due to gravity is :
$$ g=G \times \frac{M}{R^{2}} $$

Here, Gravitational constant :

$G=6.7 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$

Mass of the moon, $M=7.4 \times 10^{22} \mathrm{~kg}$

Radius of the moon, $R=1740 \mathrm{~km}$
$$
\begin{array}{l}

=1740 \times 1000 \mathrm{~m} \\

=1.74 \times 10^{6} \mathrm{~m}
\end{array} $$

Now, putting these values of $G, M$ and $R$ in the above formula, we get:
$$ g=\frac{6.7 \times 10^{-11} \times 7.4 \times 10^{22}}{\left(1.74 \times 10^{6}\right)^{2}} $$

$$
\text { or } \quad g=1.63 \mathrm{~m} / \mathrm{s}^{2}
$$

Thus, the acceleration due to gravity, $g$, on the surface of the moon is $1.63 \mathrm{~m} / \mathrm{s}$

by
7.2k Points

Related Questions

1 Vote
1 Vote
1 Answer 15 Views
3 Votes
3 Votes
0 Answers 25 Views
1 Vote
1 Vote
1 Answer 12 Views

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]

Connect us on Social Media