To determine the radius of a planet given its mass and surface gravity, we can use the universal law of gravitation. Here’s a step-by-step solution to solving this problem:
1. Identify the given values and the required formula:
- The mass of the planet, [tex]\( M = 3.40 \times 10^{23} \, \text{kg} \)[/tex]
- The surface gravity, [tex]\( g = 6.50 \, \text{m/s}^2 \)[/tex]
- The gravitational constant, [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
2. Recognize the formula for gravitational force (weight):
[tex]\[
g = \frac{G \cdot M}{r^2}
\][/tex]
where:
- [tex]\( g \)[/tex] is the surface gravity,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the planet,
- [tex]\( r \)[/tex] is the radius of the planet.
3. Rearrange the formula to solve for the radius [tex]\( r \)[/tex]:
Start by isolating [tex]\( r^2 \)[/tex]:
[tex]\[
r^2 = \frac{G \cdot M}{g}
\][/tex]
Then take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[
r = \sqrt{\frac{G \cdot M}{g}}
\][/tex]
4. Substitute the known values into the equation:
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
- [tex]\( M = 3.40 \times 10^{23} \, \text{kg} \)[/tex]
- [tex]\( g = 6.50 \, \text{m/s}^2 \)[/tex]
5. Calculate the value inside the square root:
[tex]\[
\frac{G \cdot M}{g} = \frac{6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 3.40 \times 10^{23} \, \text{kg}}{6.50 \, \text{m/s}^2}
\][/tex]
[tex]\[
\frac{6.67430 \times 10^{-11} \times 3.40 \times 10^{23}}{6.50} = 3.4905761538461538 \times 10^{12} \, \text{m}^2
\][/tex]
6. Take the square root to find the radius [tex]\( r \)[/tex]:
[tex]\[
r = \sqrt{3.4905761538461538 \times 10^{12}} \approx 1868467.9038432282 \, \text{m}
\][/tex]
Thus, the radius of the planet is approximately [tex]\( 1,868,467.90 \, \text{m} \)[/tex], or about [tex]\( 1.87 \times 10^6 \, \text{m} \)[/tex].
