Question

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10^24 kg and radius of the earth is 6.4 × 10^6 m).​

Answer 1

Question:

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10^24 kg and radius of the earth is 6.4 × 10^6 m).

Solution:

${\underline {\underline {\boxed {\rm {\green { Given }}}}}}$

• $\sf\blue { m_1 }$ ( mass of the earth ) = $\sf\ { 6 \times {10}^{24} kg }$

• $\sf\blue { m_2 }$ ( mass of the object ) = $\sf\ { 1kg }$

• $\sf\blue { G }$ ( universal gravitation constant ) = $\sf\ { 6.7 \times {10}^{-11} N{m}^{2} / k{g}^{2}}$

• $\sf\blue { r }$ ( radius ) = $\sf { 6.4 \times {10}^{6} m }$

${\underline {\underline {\boxed {\rm {\green { To \: find: }}}}}}$

• Magnitude of the gravitational force between the earth and a 1 kg object on its surface.

${\underline {\underline {\boxed {\rm {\green { Calculation: }}}}}}$

We know,

${\underline {\underline {\boxed {\rm {\blue { F = \dfrac{Gm_1m_2}{{r}^{2} }}}}}}}$

According to formula, substituting values-

$\sf { \implies \: F = \dfrac{6.7 \times {10}^{ - 11 } \times 6 \times {10}^{24} \times 1}{( \: 6.4 \times {10}^{6} ) {}^{2} } }$

$\:$

$\sf { \implies \: F = \dfrac{ 6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1 }{6.4 \times 6.4 \times {10}^{6} \times {10}^{6} } }$

$\:$

$\sf { \implies \: F = \dfrac{6.7 \times6 \times {10}^{( - 11 + 24)} }{6.4 \times 6.4 \times {10}^{(6 + 6)} } }$

$\:$

$\sf { \implies \: F = \dfrac{6.7 \times 6 \times {10}^{13} }{6.4 \times 6.4 \times {10}^{12} } }$

$\:$

$\sf { \implies \: F = \dfrac{40.2 }{40.96} \times {10}^{(13 - 12)} }$

$\:$

$\sf { \implies \: F = \dfrac{40.2 }{40.96} \times 10 }$

$\:$

$\sf { \implies \: F = \dfrac{402 \: \times \: \cancel{ 10 } \: \times 100}{4096 \: \times \: \cancel{10}} }$

$\:$

$\sf { \implies \: F = \cancel {\dfrac{40200}{4096} } = 9.8N}$

Therefore , magnitude of the gravitational force between the earth and a 1 kg object on its surface is 9.8 Newtons.

___________________________

Additional Information:

Points to remember while solving numericals related to gravitation:

• When a body is dropped freely from a height then, u ( intial velocity) = 0
• When a body is thrown vertically upwards then, v ( final velocity ) = 0
• When a body is falling vertically downwards then, g ( acceleration due to gravity ) = +9.8m/s²
• When a body is thrown vertically upwards then, g = -9.8m/s²

Equations of motion for freely falling bodies:

• ${\underline {\underline {\boxed {\rm {\blue { v = u + gt }}}}}}$

• ${\underline {\underline {\boxed {\rm {\blue { h = ut + \frac{1}{2}g{t}^{2} }}}}}}$

• ${\underline {\underline {\boxed {\rm {\blue { {v}^{2} = {u}^{2} + 2gh }}}}}}$

Where,

• v is final velocity
• u is intial velocity
• g is acceleration due to gravity
• h is height
• t is time

___________________________

Answer 2

Answer:

Given:

Mass of earth m₁ =6 × 10²⁴ kg

Mass of the object m₂ = 1 kg

Radius of the earth = 6.4 × 10⁶ m

Universal gravitational constant G = 6.67 x 10⁻¹¹ Nm²kg⁻²

By applying universal law of gravitation: - F = (G m₁ m₂/r²)  

= (6.67 x 10⁻¹¹ x 6 × 10²⁴ x 1)/ (6.4 × 10⁶)²

F = 9.8 N