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The radius of the earth is 6400 km and $g=10ms^{-2}$. In order that a body of 5 kg weights zero at the equator, the angular speed of the earth is = rad/sec
for condition of weight lessness at equator $ \omega = \sqrt { g / R } $ $ = \sqrt { 10 \over 6400 \times 10^3 } = { 1 \over 800} rad /sec $
The time period of a simple pendulum on a freely moving artificial satellite is
Time peripd of simple pendylym $ T = 2 \pi \sqrt { l /g } $ In artificial satellite g ' = 0 $ \therefore T = \infty $
The value of g on he earth surface is $980 cm/sec^2$. Its value at a height of 64 km from the earth surface is …………..$cm5^{–2} $
$ { g' \over g } = \left( { R \over (R+h) } \right)^2 = \left( {6400 \over 6400+64} \right) \Rightarrow g' = 960.400 ms^{-2} $
If earth rotates faster than its present speed the weight of an object will.
$ g' = g - \omega^2 R cos^2 \lambda $ Rotation of the earth results in the decreased weight apparently.this decrease in weight is not felt at the poles as the angle of latitude is 90'
The moon’s radius is 1/4 that of earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of earth, that on the surface of the moon is
using $ g = { GM \over R^2 } $we get $ g_m = g/5 $
The depth of at which the value of acceleration due to gravity becomes 1/n the time the value of at the surface is (R = radius of earth)
$ g^1 = g ({ 1 - { d\over R }} ) \Rightarrow d = { (n-1) \over n } R $
If the density of small planet is that of the same as that of the earth while the radius of the planet is 0.2 times that of l the earth, the gravitational acceleration on the surface of the planet is ...............
$ g = { 4 \over 3 } \pi GR \rho \, and, g' = {4 \over 3 } \pi GR' \rho $ $ \therefore { g' \over g } = {R' \over R } = 0.2 \Rightarrow g' = 0.2 g $
If mass of a body is M on the earth surface, than the mass of the same body on the moon surfae is
mass does not vary from place to place
If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is
$ g' = g \left( { R \over R+ h } \right)^2 = 9/4 by solving \, h = R $
If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the earth is $9.8 ms^{-2}$ , then the value of ‘g’ on that planet is = ............... $ms^{-2}$
$g_p = g_e \left( { M_p \over M_e } \right) = \left( { R_e \over R_p } \right)^2 = 9.8 \left( { 1 \over 80} \right)(2)^2 = { 9.8 \over 20 } = 0.49 ms^{-2} $
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