The mass of a planet is 1/10 th  that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is :

We need the surface gravity \(g\) of a planet whose mass and size are given relative to Earth.

The universal formula for gravitational acceleration at the surface of a spherical body is

$g\; =\; \backslash frac\{G\; M\}\{R^\{2\}\}$

where \(G\) is the gravitational constant, \(M\) is the mass, and \(R\) is the radius of the body.

For Earth we know

$g\_\{\backslash text\{Earth\}\}\; =\; 9.8\backslash \; \backslash text\{m\; s\}–2$

Since the same constant \(G\) appears for both Earth and the planet, we can form a ratio:

$\backslash frac\{g\_\{\backslash text\{planet\}\}\}\{g\_\{\backslash text\{Earth\}\}\}\; =\; \backslash frac\{M\_\{\backslash text\{planet\}\}/M\_\{\backslash text\{Earth\}\}\}\{\backslash left(R\_\{\backslash text\{planet\}\}/R\_\{\backslash text\{Earth\}\}\backslash right)^\{2\}\}$

The problem states:

• Mass of the planet = \(\frac{1}{10}\) × mass of Earth → \(M_{\text{planet}}/M_{\text{Earth}} = \frac{1}{10}\)
• Diameter of the planet = \(\frac{1}{2}\) × diameter of Earth → radius also \(\frac{1}{2}\) × Earth’s radius → \(R_{\text{planet}}/R_{\text{Earth}} = \frac{1}{2}\)

Plug these ratios into the expression:

$\backslash frac\{g\_\{\backslash text\{planet\}\}\}\{9.8\}\; =\; \backslash frac\{\backslash frac\{1\}\{10\}\}\{\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{2\}\}\; =\; \backslash frac\{\backslash frac\{1\}\{10\}\}\{\backslash frac\{1\}\{4\}\}\; =\; \backslash frac\{4\}\{10\}\; =\; 0.4$

Therefore

$g\_\{\backslash text\{planet\}\}\; =\; 0.4\; \backslash times\; 9.8\; =\; 3.92\backslash \; \backslash text\{m\; s\}–2$

Hence the acceleration due to gravity on that planet is $3.92$ m s^–2.