Question
A planet ‘P’ has a mass 2 times and radius 3 times that of the Earth. What is the acceleration due to gravity on the planet, if the acceleration due to gravity on the Earth is 10 m/s2.
Answers
GPT 3.5
The acceleration due to gravity on a planet can be determined using the formula:
\(g = \frac{{G \cdot M}}{{R^2}}\)
Where:
- g is the acceleration due to gravity
- G is the gravitational constant (approximately \(6.67430 \times 10^{-11} \, \text{m}^3/\text{kg} \, \text{s}^2\))
- M is the mass of the planet
- R is the radius of the planet
Given that the mass of planet P is 2 times and the radius is 3 times that of Earth, we can calculate the acceleration due to gravity on planet P:
Mass of planet P = 2 times the mass of Earth = \(2 \times M_{\oplus}\)
Radius of planet P = 3 times the radius of Earth = \(3 \times R_{\oplus}\)
Substituting the values in the formula, we get:
\(g_P = \frac{{G \cdot (2 \cdot M_{\oplus})}}{{(3 \cdot R_{\oplus})^2}}\)
Since the acceleration due to gravity on Earth is 10 m/s², we have:
\(10 = \frac{{G \cdot M_{\oplus}}}{{R_{\oplus}^2}}\)
Simplifying this equation, we find:
\(G \cdot M_{\oplus} = 10 \cdot R_{\oplus}^2\)
Now, substituting this value in the previous equation for planet P:
\(g_P = \frac{{10 \cdot R_{\oplus}^2 \cdot 2}}{{(3 \cdot R_{\oplus})^2}}\)
Simplifying further:
\(g_P = \frac{{20}}{{9}} \cdot g_{\oplus}\)
Therefore, the acceleration due to gravity on planet P is \( \frac{{20}}{{9}} \) times Earth's acceleration due to gravity.
Given that Earth's acceleration is 10 m/s², the acceleration due to gravity on planet P is:
\(g_P = \frac{{20}}{{9}} \cdot 10 = \frac{{200}}{{9}} \approx 22.22 \, \text{m/s²}\)
\(g = \frac{{G \cdot M}}{{R^2}}\)
Where:
- g is the acceleration due to gravity
- G is the gravitational constant (approximately \(6.67430 \times 10^{-11} \, \text{m}^3/\text{kg} \, \text{s}^2\))
- M is the mass of the planet
- R is the radius of the planet
Given that the mass of planet P is 2 times and the radius is 3 times that of Earth, we can calculate the acceleration due to gravity on planet P:
Mass of planet P = 2 times the mass of Earth = \(2 \times M_{\oplus}\)
Radius of planet P = 3 times the radius of Earth = \(3 \times R_{\oplus}\)
Substituting the values in the formula, we get:
\(g_P = \frac{{G \cdot (2 \cdot M_{\oplus})}}{{(3 \cdot R_{\oplus})^2}}\)
Since the acceleration due to gravity on Earth is 10 m/s², we have:
\(10 = \frac{{G \cdot M_{\oplus}}}{{R_{\oplus}^2}}\)
Simplifying this equation, we find:
\(G \cdot M_{\oplus} = 10 \cdot R_{\oplus}^2\)
Now, substituting this value in the previous equation for planet P:
\(g_P = \frac{{10 \cdot R_{\oplus}^2 \cdot 2}}{{(3 \cdot R_{\oplus})^2}}\)
Simplifying further:
\(g_P = \frac{{20}}{{9}} \cdot g_{\oplus}\)
Therefore, the acceleration due to gravity on planet P is \( \frac{{20}}{{9}} \) times Earth's acceleration due to gravity.
Given that Earth's acceleration is 10 m/s², the acceleration due to gravity on planet P is:
\(g_P = \frac{{20}}{{9}} \cdot 10 = \frac{{200}}{{9}} \approx 22.22 \, \text{m/s²}\)