Answer:
The acceleration due to gravity at the surface of a planet is given by the formula:
\[ g = \frac{G \cdot M}{r^2} \]
Where:
- \( g \) is the acceleration due to gravity.
- \( G \) is the gravitational constant (\(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)).
- \( M \) is the mass of the planet.
- \( r \) is the radius of the planet.
Since the planet has the same mass as the Earth, we can denote \( M \) as the mass of the Earth, and we are given that the radius of the planet is 40% greater than that of the Earth, which means \( r = 1.4 \times \text{radius of Earth} \).
Plugging these values into the formula, we get:
\[ g = \frac{G \cdot M}{(1.4 \cdot \text{radius of Earth})^2} \]
However, we know that the mass of the planet (\( M \)) is the same as the mass of the Earth, so we can cancel out the \( M \) terms:
\[ g = \frac{G}{(1.4 \cdot \text{radius of Earth})^2} \]
Now we need to consider that the radius of the Earth cancels out as well, as it's in the denominator:
\[ g = \frac{G}{1.4^2} \]
Finally, calculate the value of \( g \) using the gravitational constant and simplify:
\[ g = \frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}}{1.4^2} \approx 5.291 \times 10^{-11} \, \text{m/s}^2 \]
So, the acceleration due to gravity at the surface of the planet is approximately \( 5.291 \times 10^{-11} \, \text{m/s}^2 \).
Step-by-step explanation: