Answer :

I don't know Ioput, but :

gravity is:
[tex]F_g = \frac{mMG}{ R^{2} } [/tex]

simplified to near surface force:
[tex]F_g = mg[/tex]

so :
[tex]g = \frac{MG}{ R^{2} } [/tex]

M mass of planet
R radius of planet
G gravitational constant

The acceleration due to gravity on the surface of the planet Loput is [tex]\boxed{1.93g}[/tex] .

Further Explanation:

Given:

The mass of the Earth is [tex]{M_e}[/tex] .

The radius of the Earth is [tex]{R_e}[/tex] .

The acceleration due to gravity on Earth is [tex]g[/tex] .

The mass of the planet Loput is [tex]5.6{M_e}[/tex] .

The radius of the planet Loput is [tex]1.7{R_e}[/tex] .

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:

[tex]\boxed{{g_p}=\frac{{G{M_p}}}{{R_p^2}}}[/tex]

Here, [tex]{g_p}[/tex]  is the acceleration due to gravity on the planet, [tex]G[/tex]  is the gravitational constant, [tex]{M_p}[/tex]  is the mass of the planet and [tex]{R_p}[/tex]  is the radius of the planet.

The value of acceleration due to gravity on the surface of the Earth is:

[tex]g=\frac{{G{M_e}}}{{R_e^2}}[/tex]

Now for the planet Loput, the value of acceleration due to gravity can be determined as:

[tex]\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}[/tex]

Substitute [tex]g[/tex]  for  [tex]\frac{{G{M_e}}}{{{R_e}}}[/tex] in above expression.

[tex]\boxed{{g_L}=1.93g}[/tex]

Thus, the acceleration due to gravity on the surface of the planet Loput is [tex]\boxed{1.93g}[/tex] .

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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.

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