Mass of the earth is 81 times the mass of the moon and distance between the earth and moon is 60 times the radius of the earth. If R is radius of the earth, then the distance between moon and the point on the line joining the moon and the earth where the gravitation force becomes zero is

30R

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15R

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Solution

The correct option is C 6R
Let m be the mass at a distance x from the centre of the moon where gravitational force is zero.
$∴ \frac{{G M}_{e} m}{(60 R - x)^{2}} = \frac{{G M}_{m o o n} m}{x^{2}} o r \frac{81}{(60 R - x)^{2}} = \frac{1}{x^{2}} o r \frac{9}{60 R - x} = \frac{1}{x} o r x = 6 R$

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Mass of the earth is 81 times the mass of the moon and distance between the earth and moon is 60 times the radius of the earth. If R is radius of the earth, then the distance between moon and the point on the line joining the moon and the earth where the gravitation force becomes zero is

The correct option is C 6RLet m be the mass at a distance x from the centre of the moon where gravitational force is zero.∴GMem(60R−x)2=GMmoonmx2or81(60R−x)2=1x2or960R−x=1xorx=6R

The correct option is C 6R
Let m be the mass at a distance x from the centre of the moon where gravitational force is zero.
$∴ \frac{{G M}_{e} m}{(60 R - x)^{2}} = \frac{{G M}_{m o o n} m}{x^{2}} o r \frac{81}{(60 R - x)^{2}} = \frac{1}{x^{2}} o r \frac{9}{60 R - x} = \frac{1}{x} o r x = 6 R$