The earth's gravitational field intensity at its surface is about (G = 6.7 × 10 ^ -11 Nm ^2 /kg ^2 , mass of…

Analysis

The question asks for the gravitational field intensity ($g$) at the surface of the Earth. Gravitational field intensity is defined as the force per unit mass exerted by a gravitational field.

Concept

The formula for gravitational field intensity ($g$) at the surface of a planet is derived from Newton's Law of Universal Gravitation:

Where:

• $G$ = Universal gravitational constant ($6.7 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$)
• $M$ = Mass of the Earth ($6 \times 10^{24} \, \text{kg}$)
• $R$ = Radius of the Earth ($6.4 \times 10^6 \, \text{m}$)

Solution

Let's plug the given values into the formula:

1. Calculate the numerator ($G \times M$): $G \times M = (6.7 \times 10^{-11}) \times (6 \times 10^{24})$ $G \times M = 40.2 \times 10^{13}$ $G \times M = 4.02 \times 10^{14}$

2. Calculate the denominator ($R^2$): $R^2 = (6.4 \times 10^6)^2$ $R^2 = 40.96 \times 10^{12}$ $R^2 = 4.096 \times 10^{13}$

3. Divide the results: $g = \frac{4.02 \times 10^{14}}{4.096 \times 10^{13}}$ $g = \frac{40.2}{4.096}$ $g \approx 9.814 \, \text{N/kg}$

Rounding to one decimal place, we get 9.8 N/kg.

Conclusion

The calculation confirms that the gravitational field intensity at the Earth's surface is approximately 9.8 N/kg, which is equivalent to the acceleration due to gravity ($9.8 \, \text{m/s}^2$).

Exam Tip: Always remember that the units N/kg (Newtons per kilogram) and m/s² (meters per second squared) are equivalent for gravitational field intensity and acceleration due to gravity. If you see either unit in a question about gravity, they are referring to the same physical quantity!