The acceleration due to gravity at the surface of the moon is 1-67 m-sec-2- If the radius of the moon is 1-74 - 10-6 m- Find the mass of the moon- - Given- G - 6-67 - 10-11 Nm-2-kg-2-

G_moon = 1.67 m/sec^2 and R_moon = 1.74 × 10^6 m g = GM/R^2 So, M = gR^2/G = 1.67 × (1.74 ×10^6) ^2/6.67 ×10^ 11 = 7.58 × 10^22 kg ...

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Standard X

Physics

Satellite Launch Vehicle and Jounney to PSLV

The accelerat...

Question

The acceleration due to gravity at the surface of the moon is 1.67 $\frac{m}{{s e c}^{2}}$ . If the radius of the moon is 1.74 $\times$ $10^{6}$ m. Find the mass of the moon. ( Given: G = 6.67 $\times$ $10^{- 11}$ $\frac{N m^{2}}{{k g}^{2}}$ )

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Solution

Given,
$g_{m o o n}$ = 1.67 $\frac{m}{{s e c}^{2}}$
$R_{m o o n}$ = 1.74 $\times$ $10^{6}$ m

We know that,
$g = \frac{G M}{R^{2}}$

So, $M = \frac{g R^{2}}{G}$ = $\frac{1.67 \times ({1.74 \times 10^{6})}^{2}}{6.67 \times 10^{- 11}}$ = $7.58 \times 10^{22} k g$ .
Thus, the mass of the moon is $7.58 \times 10^{22} k g$ .

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The acceleration due to gravity at the surface of the moon is 1.67 msec2. If the radius of the moon is 1.74 × 106 m. Find the mass of the moon. ( Given: G = 6.67 × 10−11 Nm2kg2)

Given,gmoon = 1.67 msec2Rmoon = 1.74 × 106 m We know that,g=GMR2 So, M=gR2G = 1.67×(1.74×106)26.67×10−11 = 7.58×1022 kg.Thus, the mass of the moon is 7.58×1022 kg.

The acceleration due to gravity at the surface of the moon is 1.67 $\frac{m}{{s e c}^{2}}$ . If the radius of the moon is 1.74 $\times$ $10^{6}$ m. Find the mass of the moon. ( Given: G = 6.67 $\times$ $10^{- 11}$ $\frac{N m^{2}}{{k g}^{2}}$ )