Question

If earth has a mass of $5.972 \times {10^{24}}kg$ and Mars has a mass of $6.39 \times {10^{23}}kg$ and the distance between them is $225 \times {10^6}m$ with what force does earth pull on Mars?

Answer 1

In order to solve this question, we should know that every object in universe attracts with another body with a force which is known as gravitational force of attraction which was given by S. Isaac Newton, here we will use the general formula of gravitational force attraction between two bodies and then will find out the magnitude of force between earth and mars.

Formula used:
If M and M’ be the mass of two bodies and distance between them is r then gravitational force of attraction between two bodies is calculated as $F = G\dfrac{{MM'}}{{{r^2}}}$ where G is known as Gravitational constant and has a fixed value of $G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$

Complete step by step answer:
According to the question, we have given that
$M = 5.972 \times {10^{24}}kg$ mass of the earth.
$M' = 6.39 \times {10^{23}}kg$ mass of the mars.
$r = 225 \times {10^6}m$ distance between earth and mars.
Let F be the gravitational force of attraction between earth and mars the, using formula
$F = G\dfrac{{MM'}}{{{r^2}}}$ on putting the value of parameters we get,
$F = 6.67 \times {10^{ - 11}} \times \dfrac{{5.972 \times 6.39 \times {{10}^{47}}}}{{{{(225)}^2} \times {{10}^{12}}}}$
on further solving,
$F = \dfrac{{254.53 \times {{10}^{24}}}}{{50625}}$
$F = 0.00502 \times {10^{24}}N$
or
$F = 5.02 \times {10^{21}}N$
Hence, the force with which earth pulls on mars has a magnitude of $F = 5.02 \times {10^{21}}N$.

Note: It should be remembered that, while solving such questions and keeping SI unit of force as Newton, always check that given masses must be in units of Kilograms and distance between the bodies must be in metres. This gravitational force of attraction acts on every object with another object in the whole physical universe.