Answer :
Let's solve this question step-by-step.
### Given values:
- Mass of the Earth, [tex]\( M_{earth} = 6 \times 10^{24} \, \text{kg} \)[/tex]
- Radius of the Earth, [tex]\( R_{earth} = 6.4 \times 10^6 \, \text{m} \)[/tex]
- Mass of the apple, [tex]\( m_{apple} = 200 \, \text{g} = 0.2 \, \text{kg} \)[/tex] (converted to kilograms)
- Gravitational constant, [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
### i) Acceleration produced on the apple by the Earth
Newton's law of universal gravitation states that the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = G \frac{m_1 \times m_2}{r^2} \][/tex]
Here, [tex]\( m_1 = M_{earth} \)[/tex], [tex]\( m_2 = m_{apple} \)[/tex], and [tex]\( r = R_{earth} \)[/tex]. The acceleration [tex]\( a_{apple} \)[/tex] produced on the apple by the Earth is given by:
[tex]\[ a_{apple} = \frac{F}{m_{apple}} \][/tex]
Since [tex]\( F = G \frac{M_{earth} \times m_{apple}}{R_{earth}^2} \)[/tex]:
[tex]\[ a_{apple} = \frac{G M_{earth}}{R_{earth}^2} \][/tex]
After substituting the known values:
[tex]\[ a_{apple} = \frac{6.67430 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^6)^2} \][/tex]
The calculation gives us the acceleration produced on the apple by the Earth:
[tex]\[ a_{apple} \approx 9.776806640625 \, \text{m/s}^2 \][/tex]
### ii) Acceleration produced on the Earth by the apple
Using Newton’s third law of motion, the force exerted by the apple on the Earth is equal in magnitude and opposite in direction to the force exerted by the Earth on the apple.
Hence, we use the same force [tex]\( F \)[/tex], but the acceleration [tex]\( a_{earth} \)[/tex] produced on the Earth is given by:
[tex]\[ a_{earth} = \frac{F}{M_{earth}} \][/tex]
Since [tex]\( F = G \frac{M_{earth} \times m_{apple}}{R_{earth}^2} \)[/tex]:
[tex]\[ a_{earth} = \frac{G m_{apple}}{R_{earth}^2} \][/tex]
After substituting the known values:
[tex]\[ a_{earth} = \frac{6.67430 \times 10^{-11} \times 0.2}{(6.4 \times 10^6)^2} \][/tex]
The calculation gives us the acceleration produced on the Earth by the apple:
[tex]\[ a_{earth} \approx 3.258935546875 \times 10^{-25} \, \text{m/s}^2 \][/tex]
### Final Answer:
1. The acceleration produced on the apple by the Earth is approximately [tex]\( 9.776806640625 \, \text{m/s}^2 \)[/tex].
2. The acceleration produced on the Earth by the apple is approximately [tex]\( 3.258935546875 \times 10^{-25} \, \text{m/s}^2 \)[/tex].
### Given values:
- Mass of the Earth, [tex]\( M_{earth} = 6 \times 10^{24} \, \text{kg} \)[/tex]
- Radius of the Earth, [tex]\( R_{earth} = 6.4 \times 10^6 \, \text{m} \)[/tex]
- Mass of the apple, [tex]\( m_{apple} = 200 \, \text{g} = 0.2 \, \text{kg} \)[/tex] (converted to kilograms)
- Gravitational constant, [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
### i) Acceleration produced on the apple by the Earth
Newton's law of universal gravitation states that the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = G \frac{m_1 \times m_2}{r^2} \][/tex]
Here, [tex]\( m_1 = M_{earth} \)[/tex], [tex]\( m_2 = m_{apple} \)[/tex], and [tex]\( r = R_{earth} \)[/tex]. The acceleration [tex]\( a_{apple} \)[/tex] produced on the apple by the Earth is given by:
[tex]\[ a_{apple} = \frac{F}{m_{apple}} \][/tex]
Since [tex]\( F = G \frac{M_{earth} \times m_{apple}}{R_{earth}^2} \)[/tex]:
[tex]\[ a_{apple} = \frac{G M_{earth}}{R_{earth}^2} \][/tex]
After substituting the known values:
[tex]\[ a_{apple} = \frac{6.67430 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^6)^2} \][/tex]
The calculation gives us the acceleration produced on the apple by the Earth:
[tex]\[ a_{apple} \approx 9.776806640625 \, \text{m/s}^2 \][/tex]
### ii) Acceleration produced on the Earth by the apple
Using Newton’s third law of motion, the force exerted by the apple on the Earth is equal in magnitude and opposite in direction to the force exerted by the Earth on the apple.
Hence, we use the same force [tex]\( F \)[/tex], but the acceleration [tex]\( a_{earth} \)[/tex] produced on the Earth is given by:
[tex]\[ a_{earth} = \frac{F}{M_{earth}} \][/tex]
Since [tex]\( F = G \frac{M_{earth} \times m_{apple}}{R_{earth}^2} \)[/tex]:
[tex]\[ a_{earth} = \frac{G m_{apple}}{R_{earth}^2} \][/tex]
After substituting the known values:
[tex]\[ a_{earth} = \frac{6.67430 \times 10^{-11} \times 0.2}{(6.4 \times 10^6)^2} \][/tex]
The calculation gives us the acceleration produced on the Earth by the apple:
[tex]\[ a_{earth} \approx 3.258935546875 \times 10^{-25} \, \text{m/s}^2 \][/tex]
### Final Answer:
1. The acceleration produced on the apple by the Earth is approximately [tex]\( 9.776806640625 \, \text{m/s}^2 \)[/tex].
2. The acceleration produced on the Earth by the apple is approximately [tex]\( 3.258935546875 \times 10^{-25} \, \text{m/s}^2 \)[/tex].
