Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The radius of the Earth is 6370 km and the radius of mars is 3440 km. What is the acceleration due to Gravity on Mars if the mass of Mars is $1.1$ times the mass of Earth?

Last updated date: 19th Apr 2024
Total views: 35.1k
Views today: 0.35k
Verified
35.1k+ views
Hint: In this solution, we will use the formula of gravitational acceleration on a planet. We will find the gravitational acceleration on the surface of Earth and Mass and take their ratio to determine the acceleration due to gravity on the surface of Mars.
Formula used: In this solution, we will use the following formulae:
Gravitational acceleration on any planet: $g = \dfrac{{GM}}{{{R^2}}}$ where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$is the radius of the planet.

We’ve been given that the radius of the Earth is 6370 km and the radius of mars is 3440 km and the mass of Mars is $1.1$ times the mass of Earth.
Let us denote the mass and radius of Earth as ${M_e}\,{\text{and}}\,{{\text{R}}_e}$ and the mass and radius of Mars as ${M_m}\,{\text{and}}\,{{\text{R}}_m}$.
Then the gravitational acceleration on the surface of Earth will be
${g_e} = \dfrac{{G{M_e}}}{{{{\left( {6370} \right)}^2}}}$
And the gravitational acceleration on the surface of Mars will be
${g_m} = \dfrac{{G{M_m}}}{{{{\left( {3440} \right)}^2}}}$
Taking the ratio of the two gravitational acceleration, we get
$\dfrac{{{g_e}}}{{{g_m}}} = \dfrac{{{M_e}}}{{{M_m}}} \times {\left( {\dfrac{{3440}}{{6370}}} \right)^2}$
Now we’ve been given that the mass of Mars is $1.1$ times the mass of Earth so we can write that mathematically as ${M_m} = 1.1{M_e}$. So, the above equation will be transformed as
$\dfrac{{{g_e}}}{{{g_m}}} = \dfrac{1}{{1.1}} \times {\left( {\dfrac{{3440}}{{6370}}} \right)^2}$
Which gives us
$\dfrac{{{g_e}}}{{{g_m}}} = 0.265$
Now we know that the gravitational acceleration on the surface of Earth is ${g_e} = 9.81\,m/{s^2}$ so we can find the gravitational acceleration on the surface of Mars as
${g_m} = \dfrac{{9.81}}{{0.265}} = 37\,m/{s^2}$

Note: The dimensions of Mars are different in reality than what is actually mentioned in the question. But this question tests the concepts of gravitational acceleration and its dependence on the mass and radius of a planet.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}