Answer: Below
Explanation:
To find the gravitational field strength \( g \) at Pluto's surface, you can use the formula:
\[ g = \frac{G \cdot M}{r^2} \]
Where:
- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
- \( M \) is the mass of Pluto, given as \( 12 \times 10^{21} \, \text{kg} \) (expressed as \( 1.2 \times 10^{22} \, \text{kg} \)).
- \( r \) is the radius of Pluto, given as \( 1,172,000 \, \text{m} \).
Plugging in the values:
\[ g = \frac{(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (1.2 \times 10^{22} \, \text{kg})}{(1,172,000 \, \text{m})^2} \]
\[ g = \frac{(8.004 \times 10^{11} \, \text{N m}^2/\text{kg}^2)}{(1.3732 \times 10^{12} \, \text{m}^2)} \]
\[ g = 0.5829 \times 10^{-1} \, \text{N/kg} \]
So, in the form of \( a.bc \times 10^{-d} \), the gravitational field strength at Pluto's surface is approximately \( 5.829 \times 10^{-2} \, \text{N/kg} \).