Question

A body weighs $36kg$ on the surface of the Earth. How much would it weigh on the surface of a planet, whose mass is $\dfrac{1}{9}$ and radius $\dfrac{1}{3}$ of that of earth?

Answer 1

Every object on the universe is under the force of the gravitational force. Any two objects attract each other and are directly proportional to the product of the mass and inversely proportional to the square of distance between the two. Here use the formula - $g = G\dfrac{M}{{{R^2}}}$ where g is the gravity of acceleration and G is the gravitational force.

Complete step by step answer:
Given that: Weight of the body on the surface of the Earth $= mg = 36kg$
Therefore, the mass of the body is $= m = 36kg$
We know that, gravity on Earth is
$g = G\dfrac{M}{{{R^2}}}$ ....... (a)
Now, when Mass is $\dfrac{1}{9}$ and radius is $\dfrac{1}{3}$ of that of the earth.
Therefore, replace –
$M{\text{ with }}\dfrac{M}{9}{\text{ and R with }}\dfrac{R}{3}$
Now, g on the surface of the planet is
${g_{planet}} = \dfrac{{G\left( {\dfrac{M}{9}} \right)}}{{{{\left( {\dfrac{R}{3}} \right)}^2}}}$
Simplify the above equation –
${g_{planet}} = \dfrac{{G\left( {\dfrac{M}{9}} \right)}}{{\left( {\dfrac{{{R^2}}}{9}} \right)}}$
Numerator’s denominator and the denominator’s denominator are the same, so they cancel each other.
${g_{planet}} = \dfrac{{GM}}{{{R^2}}}$ ....... (b)
From the equations (a) and (b), we can say that weights on the earth and on any other planet remain the same.

Note:
Remember the difference between the g (gravitational acceleration) and the G (Gravitational constant). Since in G only magnitude is important it is the scalar quantity whereas, in g both the magnitude and the direction are important, and therefore it is vector quantity. The value of “G” remains constant throughout the globe whereas, the value of “g” changes from every place on the planet. Go through certain basic parameters and the physical quantities to solve these types of word problems.