Question

Obtain the relation between the acceleration due to the gravity on the earth and acceleration due to gravity on the moon in numerical value​

Answer 1

☆ To Find :

The Relation between the Acceleration due to gravity on the Surface of Earth and on the Surface of Moon.

☆ We Know :

$\bullet$ Acceleration due to gravity :

$\purple{\sf{\underline{\boxed{g = \dfrac{G\:M}{r^{2}}}}}}$

Where ,

• g = Acceleration due to Gravity

• G = Universal Gravitational Constant

• r = Distance

• M = mass of the planet

$\bullet$ Mass of Earth :

$\sf{M_{e} = 6 \times 10^{24} kg}$

$\bullet$ Mass of Moon :

$\sf{M_{m} = 7.3 \times 10^{23} kg}$

$\bullet$ Radius of Earth :

$\sf{R_{e} = 6.4 \times 10^{6} m }$

$\bullet$ Radius of Moon :

$\sf{R_{m} = 1.74 \times 10^{6} m}$

$\bullet$ Universal Gravitational Constant :

$\sf{G = 6.67 \times 10^{-11}}$

◕ Solution :

☆ Concept :

By finding the acceleration due to gravity on the individual planets and by comparing them , we will get the relation between the Acceleration due to Gravity Surface of Earth and on the Surface of Moon.

☆ Calculation :

Acceleration due to Gravity on the Surface of Earth :

Given :

• Mass of Earth = 6 × 10²⁴ kg.

• Radius of Earth = 6.4 × 10⁶ m.

Using the formula and substituting the values in it, we get :

$\purple{\sf{g = \dfrac{G\:M}{r^{2}}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{\left(6.4 \times 10^{6}\right)^{2}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 10^6 \times 6.4 \times 10^{6}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4 \times 6.4 \times 10^{12}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{40.96 \times 10^{12}}}$

$\implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^{-12}}{40.96}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-23} \times 6 \times 10^{24}}{40.96}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 6 \times 10}{40.96}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 60}{40.96}} \\ \\ \\ \\ \implies \sf{g = \dfrac{400.2}{40.96}} \\ \\ \\ \\ \implies \sf{g = 9.77 (approx.) ms^{-2}} \\ \\ \\ \\ \therefore \purple{\sf{g = 9.8 ms^{-2}}}$

Hence , the Acceleration due to Gravity on the surface of Earth is 9.8 m/s².

Acceleration due to Gravity on the Surface of Moon :

Given :

• Mass of Moon = 7.3 × 10²² kg.

• Radius of Moon = 1.74 × 10⁶ m.

Using the formula and substituting the values in it, we get :

$\purple{\sf{g = \dfrac{G\:M}{r^{2}}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{22}}{\left(1.74 \times 10^{6}\right)^{2}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 1.73 \times 10^{22}}{1.74 \times 10^6 \times 1.74 \times 10^{6}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{22}}{1.74 \times 1.74 \times 10^{12}}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{24}}{3.02 \times 10^{12}}}$

$\implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{24} \times 10^{-12}}{40.96}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 10^{-23} \times 7.3 \times 10^{22}}{3.02}} \\ \\ \\ \\ \implies \sf{g =\dfrac{6.67 \times 7.3 \times 10^{-1}}{3.02}} \\ \\ \\ \\ \implies \sf{g = \dfrac{6.67 \times 0.73}{3.02}} \\ \\ \\ \\ \implies \sf{g = \dfrac{4.9}{3.02}} \\ \\ \\ \\ \implies \sf{g = 1.62 (approx.) ms^{-2}} \\ \\ \\ \\ \therefore \purple{\sf{g = 1.6 ms^{-2}}}$

Hence , the Acceleration due to Gravity on the surface of moon is 1.6 m/s².

By Comparing them , we get :

$\bigstar \boxed{\sf{\dfrac{g_{e}}{g_{m}}}} \\ \\ \implies \sf{\dfrac{9.8}{1.6}} \\ \\ \implies \sf{6} \\ \\ \therefore \purple{\sf{\dfrac{g_{e}}{g_{m}} = 6}}$

Hence , the Relation is Acceleration due to gravity on earth is 6 times of Acceleration due to gravity on Moon.i.e,

$\therefore \large{\purple{\sf{g_{e} = 6g_{m}}}}$

☆ Additional information :

• First Equation of Motion = v = u + gt

• Second Equation of Motion = h = ut + ½gt

• Third Equation of Motion = v² = u² + 2gh

Answer 2

☆ To Find :

The Relation between the Acceleration due to gravity on the Surface of Earth and on the Surface of Moon.

☆ We Know :

Acceleration due to gravity :

Where ,

g = Acceleration due to Gravity

G = Universal Gravitational Constant

r = Distance

M = mass of the planet

Mass of Earth :

Mass of Moon :

Radius of Earth :

Radius of Moon :

Universal Gravitational Constant :

◕ Solution :

☆ Concept :

By finding the acceleration due to gravity on the individual planets and by comparing them , we will get the relation between the Acceleration due to Gravity Surface of Earth and on the Surface of Moon.

☆ Calculation :

Acceleration due to Gravity on the Surface of Earth :

Given :

Mass of Earth = 6 × 10²⁴ kg.

Radius of Earth = 6.4 × 10⁶ m.

Using the formula and substituting the values in it, we get :

Hence , the Acceleration due to Gravity on the surface of Earth is 9.8 m/s².

Acceleration due to Gravity on the Surface of Moon :

Given :

Mass of Moon = 7.3 × 10²² kg.

Radius of Moon = 1.74 × 10⁶ m.

Using the formula and substituting the values in it, we get :

Hence , the Acceleration due to Gravity on the surface of moon is 1.6 m/s².

By Comparing them , we get :

Hence , the Relation is Acceleration due to gravity on earth is 6 times of Acceleration due to gravity on Moon.i.e,

☆ Additional information :

First Equation of Motion = v = u + gt

Second Equation of Motion = h = ut + ½gt

Third Equation of Motion = v² = u² + 2gh☆ To Find :

The Relation between the Acceleration due to gravity on the Surface of Earth and on the Surface of Moon.

☆ We Know :

Acceleration due to gravity :

Where ,

g = Acceleration due to Gravity

G = Universal Gravitational Constant

r = Distance

M = mass of the planet

Mass of Earth :

Mass of Moon :

Radius of Earth :

Radius of Moon :

Universal Gravitational Constant :

◕ Solution :

☆ Concept :

By finding the acceleration due to gravity on the individual planets and by comparing them , we will get the relation between the Acceleration due to Gravity Surface of Earth and on the Surface of Moon.

☆ Calculation :

Acceleration due to Gravity on the Surface of Earth :

Given :

Mass of Earth = 6 × 10²⁴ kg.

Radius of Earth = 6.4 × 10⁶ m.

Using the formula and substituting the values in it, we get :

Hence , the Acceleration due to Gravity on the surface of Earth is 9.8 m/s².

Acceleration due to Gravity on the Surface of Moon :

Given :

Mass of Moon = 7.3 × 10²² kg.

Radius of Moon = 1.74 × 10⁶ m.

Using the formula and substituting the values in it, we get :

Hence , the Acceleration due to Gravity on the surface of moon is 1.6 m/s².

By Comparing them , we get :

Hence , the Relation is Acceleration due to gravity on earth is 6 times of Acceleration due to gravity on Moon.i.e,

☆ Additional information :

First Equation of Motion = v = u + gt

Second Equation of Motion = h = ut + ½gt

Third Equation of Motion = v² = u² + 2gh