Answer :
To find the acceleration of an object at sea level on Earth, we use the formula for the acceleration due to gravity, which is:
[tex]\[ a = \frac{G \times M}{r^2} \][/tex]
where:
- [tex]\( a \)[/tex] is the acceleration due to gravity,
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \)[/tex],
- [tex]\( M \)[/tex] is the mass of the Earth, approximately [tex]\( 5.97 \times 10^{24} \text{ kg} \)[/tex],
- [tex]\( r \)[/tex] is the radius of the Earth, approximately [tex]\( 6.378 \times 10^6 \text{ m} \)[/tex].
Let's step through the calculation:
1. Insert the values into the formula:
[tex]\[ a = \frac{6.67430 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.378 \times 10^6)^2} \][/tex]
2. Calculate the denominator [tex]\( (r^2) \)[/tex]:
[tex]\[ (6.378 \times 10^6)^2 = 4.0689 \times 10^{13} \][/tex]
3. Calculate the numerator [tex]\( (G \times M) \)[/tex]:
[tex]\[ 6.67430 \times 10^{-11} \times 5.97 \times 10^{24} = 3.98589 \times 10^{14} \][/tex]
4. Divide the numerator by the denominator to find [tex]\( a \)[/tex]:
[tex]\[ a = \frac{3.98589 \times 10^{14}}{4.0689 \times 10^{13}} \approx 9.79 \text{ m/s}^2 \][/tex]
Therefore, the acceleration due to gravity at sea level on Earth is approximately [tex]\( 9.79 \text{ m/s}^2 \)[/tex]. The correct answer is option [tex]\( b \)[/tex] [tex]\( 9.79 \text{ m/s}^2 \)[/tex].
[tex]\[ a = \frac{G \times M}{r^2} \][/tex]
where:
- [tex]\( a \)[/tex] is the acceleration due to gravity,
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \)[/tex],
- [tex]\( M \)[/tex] is the mass of the Earth, approximately [tex]\( 5.97 \times 10^{24} \text{ kg} \)[/tex],
- [tex]\( r \)[/tex] is the radius of the Earth, approximately [tex]\( 6.378 \times 10^6 \text{ m} \)[/tex].
Let's step through the calculation:
1. Insert the values into the formula:
[tex]\[ a = \frac{6.67430 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.378 \times 10^6)^2} \][/tex]
2. Calculate the denominator [tex]\( (r^2) \)[/tex]:
[tex]\[ (6.378 \times 10^6)^2 = 4.0689 \times 10^{13} \][/tex]
3. Calculate the numerator [tex]\( (G \times M) \)[/tex]:
[tex]\[ 6.67430 \times 10^{-11} \times 5.97 \times 10^{24} = 3.98589 \times 10^{14} \][/tex]
4. Divide the numerator by the denominator to find [tex]\( a \)[/tex]:
[tex]\[ a = \frac{3.98589 \times 10^{14}}{4.0689 \times 10^{13}} \approx 9.79 \text{ m/s}^2 \][/tex]
Therefore, the acceleration due to gravity at sea level on Earth is approximately [tex]\( 9.79 \text{ m/s}^2 \)[/tex]. The correct answer is option [tex]\( b \)[/tex] [tex]\( 9.79 \text{ m/s}^2 \)[/tex].
