Question

An apple of mass 0.15 kg falls from a tree.
What is the acceleration of the apple towards
the earth? Also calculate the acceleration of the
earth towards the apple. [Given: Mass of earth
6
Radius of earth
= 6.4 x 100 m, G = 6.67 x 10- Nm² kg ?.]
Х
1024 kg,​

Answer 1

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 :

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\boxed{\sf{g = \dfrac{GM}{r^{2}}}}$

[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 :

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

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 :

$:\implies \sf{g = \dfrac{GM}{r^{2}}} \\ \\ :\implies \sf{g = \dfrac{6.67 \times 10^{-11} \times 0.15}{(6.4 \times 10^{6})^{2}}} \\ \\ :\implies \sf{g = \dfrac{100.05 \times 10^{-11 - 2}}{40.96 \times 10^{12}}} \\ \\ :\implies \sf{g = \dfrac{100.05 \times 10^{-13 - 12}}{40.96}} \\ \\ :\implies \sf{g = 2.5(approx.) \times 10^{-25}} \\ \\ \boxed{\therefore \sf{g = 2.5(approx.) \times 10^{-25}\:ms^{-2}}}$

Therefore,

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

Answer 2

Mass of earth = 6 × 10²⁴ kg

Radius of earth = 6.4 x 10⁶ m

$G = 6.67 \times 10^{-11} Nm^2 kg^{-2}$

Given that :

Apple's mass = 0.15 kg

Earth's radius = 6.4 x 10⁶ m

Earth's mass = 6 × 10²⁴ kg

Universal gravitational constant $G = 6.67 \times 10^{-11} Nm^2 kg^{-2}$

To find :

Acceleration of apple towards the earth.

Acceleration of the earth towards the apple.

Solution :

Acceleration of apple towards the earth = 9.8 m/s²

Acceleration of the earth towards the apple = 2.5 ×10-²⁵ ms-²

Using concept :

Acceleration due to the gravity.

Using formula :

Acceleration due to the gravity $= {\boxed{\bf{g = \dfrac{GM}{r^{2}}}}}$

Where,

g means Acceleration due to the gravity.

G means Universal gravitational constant.

M means Body's mass.

r means radius of the earth.

Full solution :

Case 1st

Acceleration of apple towards the earth ↝

Using the formula putting the values we get the following results ↝

${\boxed{\bf{g = \dfrac{GM}{r^{2}}}}}\\\large{\rm{g \; = \: \dfrac{6.67 \times \; 10^{-11}} \times 6 \times \; 10^{24}{(6.4 \times 10^{6})^{2}}}}\\\large{\rm{\dfrac{40.02 \times \; 10^{-11 +24}}{40.96^{12}}}}\\\large\rm{\frac{40.02 \times 10^{13}}{40.96 \times 10^{12}}}\\\large{\rm{\dfrac{100.05 \times 10^{-13-12}}{40.96}}}$

$\large\rm{\frac{40.02 \times 10}{40.96}}\\\\\large\rm{\frac{400}{40.96}}\\\\\large\rm{\frac{40.2 \times 10}{40.96}}\\\\\large\rm{9.8 \: \: m/s^{2}}$

Case 2nd

Acceleration of earth towards the apple ↝

Using the formula putting the values we get the following results ↝

${\boxed{\bf{g = \dfrac{GM}{r^{2}}}}}\\\\\large\rm{g \; = \: \dfrac{6.67 \times \; 10^{-11} \times 0.15}{(6.4 \times 10^{6})^{2}}}$

$\large\rm{\dfrac{100.05 \times \; 10^{-11 -2}}{40.96 \times 10^{12}}}\\\\\large\rm{2.5 \times 10^{-25} \: m/s^{2}}$