Question

Derive the value of g on earth.

NO COPIED ANSWER OTHERWISE U WILL BE REPORTED

Answer 1

Explanation:

As we know that g of Earth = GM/R^2

Knowing the radius of the Earth and its mass and as the value G is 6.67×10^-11..thus we can solve it and get the g value approximately as 9.8ms^-2

Hope it Helps you

I have not copied my answer from any source and written it according to my knowledge

Answer 2

Derivation of the Value of 'g' on Earth

• Acceleration due to gravity is the acceleration of a freely falling object under the absence of any external force except gravity.
• 'g' represents the acceleration due to gravity.
• It differs from one place to another.

______________________________

Let's see how the value of 'g' on Earth is derived.

Let Mass of Earth be represented as 'M'

Let Mass of the Object be represented as 'm'

Since gravitational force acts along the line joining the centers of two bodies, the distance between Earth and the object is negligible. So, the radius of the Earth would be taken as the distance between them.

So here, the distance between object and Earth is the radius of the Earth and let it be represented as 'R'.

Let Acceleration due to gravity be represented as 'g'.

From Newton's Universal Law of Gravitation,

⇒ $\sf F = G \dfrac{Mm}{R^{2}}$        - (i)

From Newton's Second Law of Motion,

⇒ F = ma

⇒ F = mg       - (ii)

Since the value of Force would be equal in both cases we can say that,

(i) = (ii)

So taking (i) and (ii),

⇒ $\sf G\dfrac{Mm}{R^{2}} = mg$

⇒ $\sf G\dfrac{M }{R^{2}} = g$

⇒ $\sf g = G\dfrac{M }{R^{2}}$

______________________________

Here,

G is the gravitational universal constant and the value of it remains the same throughout the universe. Value of G = 6.67 × 10⁻¹¹ Nm²/kg²

M is the mass of Earth as previously mentioned and its value is 6 × 10²⁴ kg

R is the radius of Earth as previously mentioned and its value is 6.4 × 10⁶ m

Using these values, let's substitute them in our equation.

⇒ $\sf g = G\dfrac{M }{R^{2}}$

⇒$\sf g = 6.67 \times 10^{-11} \times \dfrac{6\times 10^{24} }{(6.4 \times10^{6}) ^{2}}$

⇒ $\sf g = 6.67 \times 10^{-11} \times \dfrac{6\times 10^{24} }{6.4 \times10^{6}\times 6.4 \times10^{6}}$

⇒ $\sf g = \dfrac{6.67\times 6 }{6.4\times 6.4 } \times \dfrac{10^{-11} \times 10^{24} }{10^{6} \times 10^{6} }$

⇒ $\sf g = \dfrac{40.02 }{40.96 } \times \dfrac{10^{-11+24} }{10^{6+6} }$

⇒ $\sf g = 0.977 \times \dfrac{10^{13} }{10^{12} }$

⇒ $\sf g = 0.977 \times 10^{13-12}$

⇒ $\sf g = 0.977 \times 10^{1}$

⇒ $\sf g = 9.77$

⇒ $\sf g = 9.8 \ m/s^{2}$   (approx.)

Hence, derived that the value of 'g' on Earth is 9.8 m/s²

______________________________