Question

If the diameter of mars is 6760 km and mass one-tenth that of the earth. The diameter of the earth is 12742 km. If acceleration due to gravity on earth is 9.8 ms -2, the acceleration due to gravity on mars is?

Answer 1

Hint: All we mean by "acceleration due to gravity" is the acceleration that an entity acquires as a result of being dragged down by the Earth's gravitational field. It is measured in meters per second squared \[\frac{m}{{{s^2}}}\], a SI unit. As a result of its size and direction, it is classified as a vector quantity. When talking about the acceleration due to gravity, we use the sign g. For reference, at sea level, g is given as \[9.8\]\[\frac{m}{{{s^2}}}\].

Formula used:
\[g = \frac{{Gm}}{{{R^2}}}\]
Where G is the fixed constant, m is the object's mass, and R is the distance between the objects.

Complete answer:
A body is attracted to the center of the Earth by the force known as gravity. Let's say we're talking about two masses, \[{m_a}\]and\[{m_b}\]. When two bodies are subjected to a force of the same magnitude, the relationship between those two masses is as follows:
Body inertial mass (\[{m_b}\])
= \[{m_a}\]\[[\frac{{aA}}{{aB}}]\]
Given, Diameter of mars\[ = 6760{\rm{ }}km\]
Diameter of Earth\[ = {\rm{ }}12742{\rm{ }}km\]
Acceleration due to gravity on Earth\[ = 9.8{\rm{ }}ms - 2\]

By using the formula:
\[g = \frac{{GM}}{{{R^2}}} = \frac{{4GM}}{{{D^2}}}\]

We can write it as,
\[\frac{{{g_m}}}{{{g_e}}} = \frac{{{M_m}}}{{{M_e}}} \times {(\frac{{{D_e}}}{{{D_m}}})^2}\]

On substituting the value, we obtained
\[ \Rightarrow \frac{{{g_m}}}{{9.8}} = \frac{1}{{10}} \times {(\frac{{12742}}{{6760}})^2}\]

On solving the above expressions, we obtain
\[{g_m} = 3.48m/{s^2}\]

Therefore, the acceleration due to gravity on mars is \[{g_m} = 3.48m/{s^2}\].

Note: Students should keep in mind that the acceleration due to gravity changes with altitude, depth, and latitudes at the center of the planet. And also, it should be remembered that the value of g is equal to zero. One should be careful in calculating the value of acceleration due to gravity because it includes formulas which should be applied correctly in order to get the correct solution.