Question

$\sf{\underline\pink{❥︎Question:-}}$

An apple of mass 0.15kg falls from a tree. What is the acceleration of the apple towards of the earth? Also calculate the acceleration of the earth towards the apple

$\sf [Given: \: Mass \: of \: each \: of \: earth \: =6×10^24 Kg$


And Radius of earth
$\sf{=6.4×10^6m, G=6.67Nm^2 Kg^-2}$
​

Answer 1

Answer:

Answer :

Acceleration of the apple towards the earth is 9.8 m/s².

Acceleration of the apple towards the earth is 2.5 × 10⁻²⁵ m/s².

Explanation :

Given :

Mass of the apple, m = 0.15 kg

Mass of the earth, M = 6 × 10²⁴ kg

Radius of the earth, r = 6.4 × 10⁶ m

Universal gravitational constant, G = 6.67 × 10⁻¹¹ Nm² kg⁻²

To find :

Acceleration of the apple towards the earth, g = ?

Acceleration of the apple towards the earth, g = ?

Knowledge required :

Formula for acceleration due to gravity :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

[Where : g = Acceleration due to gravity, G = Universal Gravitational constant, M = Mass of the body, r = radius of the earth]

Solution :

Acceleration of the apple towards the earth :

⠀⠀By using the formula for acceleration due to gravity and substituting the values in it, we get :

Therefore,

Acceleration of the apple towards the earth, g = 9.8 m/s².

Acceleration of the earth towards the apple :

⠀⠀By using the formula for acceleration due to gravity and substituting the values in it, we get :

Therefore,

Acceleration of the earth towards the apple, g = 2.5 × 10⁻²⁵ m/s²

Explanation:

Answer 2

Given :

• Mass of the apple, m = 0.15 kg
• Mass of the earth, M = 6 × 10²⁴ kg
• Radius of the earth, r = 6.4 × 10⁶ m
• Universal gravitational constant, G = 6.67 × 10⁻¹¹ Nm² kg⁻².

To find :

• Acceleration of the apple towards the earth, g = ?
• Acceleration of the apple towards the earth, g = ?

formula Used :

• Formula for acceleration due to gravity :

⠀⠀⠀⠀⠀⠀⠀⠀

${ \underline{\boxed{\bf{ \pink{g = \dfrac{GM}{r^{2}}}}}}} \: \: \: \bigstar$

Where

• g = Acceleration due to gravity,
• G = Universal Gravitational constant,
• M = Mass of the body,
• r = radius of the earth

Solution :

★ Acceleration of the apple towards the earth :

• ⠀⠀By using the formula for acceleration due to gravity and substituting the values in it, we get :

$\begin{lgathered}:\implies \sf{g = \dfrac{GM}{r^{2}}} \\ \\ :\implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^{6})^{2}}} \\ \\ :\implies \sf{g = \dfrac{40.02 \times 10^{-11 + 24}}{40.96 \times 10^{12}}} \\ \\ :\implies \sf{g = \dfrac{40.02 \times 10^{13}}{40.96 \times 10^{12}}} \\ \\ :\implies \sf{g = \dfrac{40.02 \times 10^{13 - 12}}{40.96}} \\ \\ :\implies \sf{g = \dfrac{40.02 \times 10}{40.96}} \\ \\ :\implies \sf{g = \dfrac{400}{40.96}} \\ \\ :\implies \sf{g = \dfrac{40.02 \times 10}{40.96}} \\ \\ :\implies \sf{g = 9.8 (approx.)}\\ \\ \end{lgathered}$

$\:\underline{\boxed{\frak{\pmb{\red{g = 9.8\:ms^{-2}}}}}}}$

Therefore,

• Acceleration of the apple towards the earth, g = 9.8 m/s².

★ Acceleration of the earth towards the apple :

• ⠀⠀By using the formula for acceleration due to gravity and substituting the values in it, we get :

$\begin{lgathered}:\implies \tt{g = \dfrac{GM}{r^{2}}} \\ \\ :\implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 0.15}{(6.4 \times 10^{6})^{2}}} \\ \\ :\implies \sf{g = \dfrac{100.05 \times 10^{-11 - 2}}{40.96 \times 10^{12}}} \\ \\ :\implies \sf{g = \dfrac{100.05 \times 10^{-13 - 12}}{40.96}} \\ \\ :\implies \sf{g = 2.5(approx.) \times 10^{-25}} \\ \\ \:\underline{\boxed{\frak{\pm{\red{g = 2.5(approx.) \times 10^{-25}\:ms^{-2}}}}}}\\ \\ \end{lgathered}$

Therefore,

• Acceleration of the earth towards the apple, g = 2.5 × 10⁻²⁵ m/s²