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The ratio g("(earth)")//g("(moon)") is e...

The ratio `g_("(earth)")//g_("(moon)")` is equal to ………

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6(approximately).
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A simple pendulum designed on the moon as a seconds pendulum is taken to a planet where the acceleration due to gravity on the surface is twice that on the Earth . If g_("earth") : g_("moon") =6 :1 find the period of oscillation of the pendulum on the planet mentionedabove.

The value of the acceleration due to gravity is g_(1) at a height h=(R)/(2) (R=radius of the earth) from the surface of the earth.It is equal to 2g_(1) at a depth d below the surface of the earth.The ratio ((d)/(R)) is equal to:

Knowledge Check

  • The ratio of g_("moon")"to" g_("earth") is

    A
    6
    B
    `(1)/(6)`
    C
    4
    D
    `(1)/(4)`

  • If g_(e) is acceleration due to gravity on earth and g_(m) is acceleration due to gravity on moon, then

    A
    `g_(e)=g_(m)`
    B
    `g_e lt g_(m)`
    C
    `g_(e)=(1)/(6)g_(e)`
    D
    `g_(m)=(1)/(6)g_(e)`

  • The weight of an object on the moon is equal to "________" of its weight on the earth.

    A
    `( 1)/( 6^(th))`
    B
    `( 1)/(8^(th))`
    C
    `( 1)/( 4^(th))`
    D
    `( 1)/( 10^(th))`

    • Similar Questions

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    The Sun and Earth each exert a gravitational force on the Moon. What is the ratio F_("sun")//F_("Earth") of these two forces? (The average Sun-Moon distance is equal to the Sun Earth distance.)

    The weight of an object on the moon is equal to of its weight on the earth.

    Masses of the earth and the moon are in the ratio 3:2 and the radii of the earth and the moon are in the ratio of 6:1 . The ratio of the weight of the body on their surface will be

    The acceleration due to gravity on the moon is only one sixth that of earth.It the earth and moon are assume to have the same density, the ratio of the radii of moon and earth will be

    If G_(1),G_(2) are the geometric means fo two series of observations and G is the GM of the ratios of the corresponding observations then G is equal to

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