Verified Answer :
I don't know Ioput, but :
gravity is:
\(F_g = \frac{mMG}{ R^{2} }\)
simplified to near surface force:
\(F_g = mg\)
so :
\(g = \frac{MG}{ R^{2} }\)
M mass of planet
R radius of planet
G gravitational constant
The acceleration due to gravity on the surface of the planet Loput is \(\boxed{1.93g}\) .
Further Explanation:
Given:
The mass of the Earth is \({M_e}\) .
The radius of the Earth is \({R_e}\) .
The acceleration due to gravity on Earth is \(g\) .
The mass of the planet Loput is \(5.6{M_e}\) .
The radius of the planet Loput is \(1.7{R_e}\) .
Concept:
In the given question, it is considered that Loput is a planet that has the mass and radius more than that of the Earth.
The value of acceleration due to gravity on the surface of any planet is expressed as:
\(\boxed{{g_p}=\frac{{G{M_p}}}{{R_p^2}}}\)
Here, \({g_p}\) is the acceleration due to gravity on the planet, \(G\) is the gravitational constant, \({M_p}\) is the mass of the planet and \({R_p}\) is the radius of the planet.
The value of acceleration due to gravity on the surface of the Earth is:
\(g=\frac{{G{M_e}}}{{R_e^2}}\)
Now for the planet Loput, the value of acceleration due to gravity can be determined as:
\(\begin{aligned}{g_L}&=\frac{{G\times\left({5.6{M_e}}\right)}}{{{{\left({1.7{R_e}}\right)}^2}}}\\&=\frac{{5.6}}{{{{\left({1.7}\right)}^2}}}\frac{{G{M_e}}}{{{R_e}}}\\\end{aligned}\)
Substitute \(g\) for \(\frac{{G{M_e}}}{{{R_e}}}\) in above expression.
\(\boxed{{g_L}=1.93g}\)
Thus, the acceleration due to gravity on the surface of the planet Loput is \(\boxed{1.93g}\) .
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Gravitation
Keywords:
Acceleration due to gravity, gravity on Loput, mass of planet Loput, 5.6Me, radius of planet Loput, 1.7Re, g=GM/R^2.