Question

If acceleration due to gravity on the earth is g, which formula gives the acceleration due to gravity on loput?

Answer 1

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 $\boxed{1.93g}$ .

Learn More:

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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.

Answer 2

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