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.
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:
\(g_P = \frac{{20}}{{9}} \cdot 10 = \frac{{200}}{{9}} \approx 22.22 \, \text{m/s²}\)
To find the acceleration due to gravity on planet P, we can use the formula:
\(g_p = \frac{{G \times M_p}}{{R_p^2}}\)
where \(g_p\) is the acceleration due to gravity on planet P, G is the gravitational constant, \(M_p\) is the mass of planet P, and \(R_p\) is the radius of planet P.
Given that the mass of planet P is 2 times that of Earth (which we know to be \(M_e\)) and the radius of planet P is 3 times that of Earth (which we know to be \(R_e\)), we can write:
\(M_p = 2 \times M_e\) and \(R_p = 3 \times R_e\)
Substituting these values into the formula, we get:
\(g_p = \frac{{G \times (2 \times M_e)}}{{(3 \times R_e)^2}}\)
Simplifying further:
\(g_p = \frac{{2 \times G \times M_e}}{{9 \times R_e^2}}\)
Since we know that the acceleration due to gravity on Earth is 10 m/s², we can substitute this value into the equation:
\(10 = \frac{{2 \times G \times M_e}}{{9 \times R_e^2}}\)
Now, we can solve for \(G \times M_e\) by multiplying both sides of the equation by \(\frac{{9 \times R_e^2}}{{2}}\):
\(G \times M_e = \frac{{10 \times 9 \times R_e^2}}{{2}}\)
Finally, we substitute this value back into the original equation for \(g_p\):
\(g_p = \frac{{2 \times (10 \times 9 \times R_e^2)}}{{9 \times R_e^2}}\)
Simplifying further:
\(g_p = 20\,m/s^2\)
Therefore, the acceleration due to gravity on planet P is \(20\,m/s^2\).