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

Answer:

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Step-by-step explanation:

Solution of your question is given in the attachment.

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Now, let's see the related topics ;

\bf Gravitational \: forceGravitationalforce : The force if attraction between two particles towards the centre of the earth, is called gravitational force.

\bf Universal \: Law \: of \: GravitationUniversalLawofGravitation : The gravitational force of attraction between any two particles is directly proportional to the product of the masses of the particles and is inversely proportional to the square of the distance between the particles.

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Answer 2

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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 ) =$\sf\ { 6.7 \times {10}^{-11} N{m}^{2} / k{g}^{2}}$

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

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

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

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

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.

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$\text{\large\underline{\red{Additional information:-}}}$

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

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