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. use the known value of G.
Answers
Answer:
Hi mate.
Thanks Thanks for asking this question.
Here is your answer,
Given :
\begin{gathered}g \: = \: 1.67 \frac{m}{s} \\ \\ r = \: 1.74 \times {10}^{6} m \\ \\ \end{gathered}
g=1.67
s
m
r=1.74×10
6
m
\begin{gathered} {g}^{o} = universal \: gravitational \: \\ \: \: \: \: \: \: \: constant \: = 6.673 \times {10}^{ - 11} \end{gathered}
g
o
=universalgravitational
constant=6.673×10
−11
m = mass of moon
We know that,
g = \frac{ {g}^{o}m }{ {r}^{2} }g=
r
2
g
o
m
1.67 = \frac{6.67 \times {10}^{ - 11} \times m }{ {(1.74 \times {10}^{6)} }^{2} }1.67=
(1.74×10
6)
2
6.67×10
−11
×m
m = \frac{1.67 \times {(1.74 \times }^{2} { {10}^{6}) }^{2} }{6.673 \times {10}^{ - 11} }m=
6.673×10
−11
1.67×(1.74×
2
10
6
)
2
m = \frac{1.67 \times 3.0276 \times {10}^{12} }{6.673 \times {10}^{ - 11} }m=
6.673×10
−11
1.67×3.0276×10
12
m = \frac{5.056 \times {10}^{23} }{6.673}m=
6.673
5.056×10
23
m = 0.7576 \times {10}^{23}m=0.7576×10
23
m = 7.576 \times {10}^{22} \: kgm=7.576×10
22
kg
The required mass of moon is
7.576 \times {10}^{22} kg7.576×10
22
kg
Thank you.