Question

Show that the acceleration due to gravity at the surface of the moon is about one-sixth of that at the surface of the earth.
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Answer 1

Answer:

Explanation:

Mass of Moon = 7.34 × 10²² kg

Radius of moon = 1737 km = 1.737 × 10⁶ m

a = GM / R²

= (6.67 × 10⁻¹¹ × 7.34 × 10²²) / (1.737 × 10⁶)²

= 1.6 m/s²

We know that acceleration due to Earth’s gravity (g) is 9.8 m/s²

Comparing a and g

a = g/6

Answer 2

Answer:

Answer :---

The explanation has been given below. Hope it is helpful.

Explanation:

Using Newton's law of gravitation we can write the formula to calculate the acceleration due to gravity on the surface of the earth.

Hence

$g_{e} = G\frac{M_{e} }{R_{e} ^{\:\:2} }$......eq(1)

In a similar way,

The acceleration due to gravity of moon :--

$g_{m} = G\frac{M_{m} }{R_{m} ^{\:\:\:2} }$.....eq(2)

On dividing eq(2) by eq(1) we will get the desired result that the ratio of $g_{e} \:and \:g_{m}$ will come out as $\frac{1}{6}$.

Mass of Moon = 7.34 × 10²² kg

Radius of moon = 1737 km = 1.737 × 10⁶ m

a = GM / R²

  = $\frac{(6.67 * 10^{-11}\: * \:7.34 * 10^{22} )}{y}$

  = 1.6 m/s²

(here (*) is (×) symbol)...

We know that acceleration due to Earth’s gravity (g) is 9.8 m/s²

Comparing a and g

a = g/6

a=$\frac{1}{6} g$