At what distance from the centre of the Moon is the point at which the strength of the resultant of the Earth's and Moon's gravitational fields is equal to zero? The Earth's mass is assumed to be $η = 81$ times that of the Moon, and the distance between the centres of these planets $n = 60$ times greater than the radius of the Earth $R$ .

Open in App

Solution

Let $r$ be the sought distance, then
$\frac{γ η M}{{(n R - r)}^{2}} = \frac{γ M}{r^{2}}$ or $η r^{2} = {(n R - r)}^{2}$
or $\sqrt{η} r = (n R - r)$ or $r = \frac{n R}{\sqrt{η} + 1} = 3.8 \times 10^{4} k m$ .

Suggest Corrections

Similar questions

Q. If the distance between the centres of earth and moon is

D

and mass of earth is

81

times that of moon. At what distance from the centre of earth gravitational field will be zero.

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.

Q. If the distance between centres of the earth and the moon is

d

and the mass of the earth is

81

times the mass of the moon, then at what distance from centre of the earth, the gravitational field will be zero?

Q. The distance between the centres of the Moon and the earth is

D

. The mass of the earth is

81

times the mass of the Moon. At what distance from the centre of the earth, the gravitational force will be zero?

Join BYJU'S Learning Program

Let r be the sought distance, then γηM(nR−r)2=γMr2 or ηr2=(nR−r)2or √ηr=(nR−r) or r=nR√η+1=3.8×104km.

Let $r$ be the sought distance, then
$\frac{γ η M}{{(n R - r)}^{2}} = \frac{γ M}{r^{2}}$ or $η r^{2} = {(n R - r)}^{2}$
or $\sqrt{η} r = (n R - r)$ or $r = \frac{n R}{\sqrt{η} + 1} = 3.8 \times 10^{4} k m$ .