Question

The acceleration due to gravity of moons surface is 1.67ms-2.If the radius of the moon is 1.74×10^6 caluculate the mass of the moon​

Answer 1

Answer:

As, g=GMR2orM=gR2Gg=GMR2orM=gR2G

This relation is true not only to the earth but for any heavenly body which is assumed to be spherical.

Now, g=1.67ms−2,R=1.74×106mg=1.67ms-2,R=1.74×106m

G=6.67×10−11Nm−2kg−2G=6.67×10-11Nm-2kg-2

∴ Mass of the moon, M=1.67×(1.74×106)26.67×10−11kgM=1.67×(1.74×106)26.67×10-11k

=7.58×1022kg=7.58×1022kg.

Answer 2

Given:-

• Acceleration due to gravity of moon's surface is 1.67m/s²
• Radius of the moon = 1.74×10^6 m

To find:-

• We've to calculate the mass of the moon=m

Solution :-

We know that,

g° = 6.673 × 10^-11

and g = g°m/r²

$\sf \implies\: 1.67ms {}^{2} = \dfrac{6.67 \times 10 {}^{ - 11} }{(1.74 \times 10 {}^{6} ) {}^{2} }$

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

$\sf \implies \: m = \dfrac{1.67 \times 3.0276 \times 10 {}^{10} }{6.673 \times 10 {}^{ - 11} }$

$\sf \implies \: m = \dfrac{5.056 \times 10 {}^{23} }{6.673}$

$\sf \implies \: 10 {}^{23} \times 0.7576$

$\sf \implies \: m = 7576 \times 10 {}^{19}$

∴ The mass of the moon is 7576 × 10^19.