Question

The distance of a point from the earth's centre where the resultant gravitational field due to the earth and the moon is zero is...( The mass of the earth is 6.0$\times$ ${10}^{24}$ kg and that of the moon is 7.4$\times$ ${10}^{22}$ kg. The distance between the earth and the moon is 40 $\times$ ${10}^5$ km.

• A. $1.6\times {10}^{5}km$
• B. $3.6\times {10}^{5}km$
• C. $4.6\times {10}^{5}km$
• D. $6\times {10}^{5}km$

Answer 1

The correct option is B

$3.6\times {10}^{5}km$

The point must be on the line joining the centres of the earth and the moon and in between them. If the distance of the point from the earth is x, the distance from the moon is (4.0$\times$${10}^5$km - x). The magnitude of the gravitational field due to the earth is

$E_1=\frac{G{M}_e}{x^2}=\frac{G\times 6\times {10}^{24kg}}{x^2}$

and magnitude of the gravitational field due to the moon is

$E_2=\frac{G{M}_m}{{(4.0\times {10}^{5}km-x)}^2}=\frac{G\times 7.4\times {10}^{22kg}}{{(4.0\times {10}^{5}km-x)}^2}$

These feild are in opposite directions. For the resultant field to be zero $E_1=E_2$

or, $\frac{6\times {10}^{24}kg}{x^2}$ = $\frac{7.4\times {10}^{22}kg}{(4.0\times {10}^{5}km-x^2{)}^2}$

or, $\frac{x}{4.0\times {10}^{5}km-x}$ = $\sqrt{\left(\frac{6\times {10}^{24}}{7.4\times {10}^{22}}\right)}$=9

or, x= 3.6 $\times {10}^5$ km