Question

The radius of the Earth is 6400 km and that of Mars is 3200 km. The mass of the Earth is about 10 times the mass of Mars. An object weighs 200N on the surface of the Earth. Its weight on the surface of mars will be:
A. 20N
B. 8N
C. 80N
D. 40N

Answer 1

Hint: Use the equation to find the weight of an object i.e. W=mg. Find the ratio between the weights of Earth and Mars respectively, \[\dfrac{{{W}_{E}}}{{{W}_{M}}}\]. to get the final answer for a given problem.

Complete Step-by-Step solution:
Weight acting on an object is nothing but the action of force of gravity on it. The product of the mass of the object and the acceleration due to gravity will give its weight. So we can say that, it completely depends on the acceleration due to gravity. On the surface of the Earth, the g is approximately equal to \[9.8{m}/{{{s}^{2}}}\;\], whereas it is not the same everywhere in this universe.
Here we have to find the weight acting on the given object on Mars's surface. Let us first write the equations to find the weight of an object on the Earth and Mars respectively.
The weight of the object on the Earth, \[{{W}_{E}}=m{{g}_{E}}\] …………..(1)
The weight of the object on the Mars, \[{{W}_{M}}=m{{g}_{M}}\]……………(2)
Where m - the mass of the object
            \[{{g}_{E}}\] - the acceleration due to gravity of the Earth
            \[{{g}_{M}}\] - the acceleration due to gravity of Mars.
The given data are
The radius of the Earth, \[{{R}_{E}}=6400km\]
The radius of the Mars, \[{{R}_{M}}=3200km\]
The mass of the Earth, \[{{M}_{E}}\]= 10\[{{M}_{M}}\] ( \[{{M}_{M}}\] - The mass of the Mars )
Now, dividing equation (1) by equation (2) i.e.
\[\dfrac{{{W}_{E}}}{{{W}_{M}}}=\dfrac{m{{g}_{E}}}{m{{g}_{M}}}\]
\[\dfrac{{{W}_{E}}}{{{W}_{M}}}=\dfrac{{{g}_{E}}}{{{g}_{M}}}\]……….....(3)
But the acceleration due to gravity is given by the equation
\[g=\dfrac{GM}{{{R}^{2}}}\]………………(4)
Using equation (4) we write \[{{g}_{E}}\] and \[{{g}_{M}}\] as
\[{{g}_{E}}=\dfrac{G{{M}_{E}}}{{{R}_{E}}^{2}}\]……………(5)
\[{{g}_{M}}=\dfrac{G{{M}_{M}}}{{{R}_{M}}^{2}}\]……………(6)
Where G is the gravitational constant. Its value is \[6.674\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\] in SI unit.
Now let us substitute the values of \[{{g}_{E}}\] and\[{{g}_{M}}\] in equation (3). Then
\[\dfrac{{{W}_{E}}}{{{W}_{M}}}=\dfrac{{}^{G{{M}_{E}}}/{}_{R_{E}^{2}}}{{}^{G{{M}_{M}}}/{}_{R_{M}^{2}}}\]
\[\dfrac{{{W}_{E}}}{{{W}_{M}}}=\dfrac{{{M}_{E}}}{{{M}_{M}}}\times \dfrac{R_{M}^{2}}{R_{E}^{2}}\]……………..(7)
But. \[{{M}_{E}}=10{{M}_{M}}\] Therefore equation (7) become
\[\dfrac{{{W}_{E}}}{{{W}_{M}}}=\dfrac{10{{M}_{M}}}{{{M}_{M}}}\times \dfrac{R_{M}^{2}}{R_{E}^{2}}\]
\[\dfrac{{{W}_{E}}}{{{W}_{M}}}=10\times \dfrac{R_{M}^{2}}{R_{E}^{2}}\]…………….(8)
Substituting the given values in equation (8)
\[\dfrac{200}{{{W}_{M}}}=10\times {{\left( \dfrac{3200}{6400} \right)}^{2}}\]
\[\dfrac{200}{{{W}_{M}}}=10\times {{\left( \dfrac{1}{2} \right)}^{2}}\]
\[{{W}_{M}}=.4\times 200N\]
\[{{W}_{M}}=80N\].
Thus the weight of the object on the Mars surface is 80N.
Therefore option C is the answer.

Note: Remember the acceleration due to gravity does not depend on the mass of the object. It depends on the mass and radius of that body which will exert the force. Since the Earth is heavier than Mars, so the object will weigh more on the surface of the Earth.