
State Newton's law of gravitation.And explain the variation of acceleration due to gravity on the surface of earth.
Answer
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Hint: Acceleration due to gravity is dependent on the distance from the center of earth and the latitude of the position.
Complete answer:
Newton's law of gravitation states that two objects having masses attract each other by a force proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
where G is the universal gravitation constant. Considering the mass of earth to be $M_e$, and its radius be $R_e$, the acceleration due to gravity g at the surface will be given by,
$g=\dfrac{G{{M}_{e}}}{R_{e}^{2}}$
Now, there are other factors which influence the acceleration due gravity. One of the factors is the latitude of the position. Since the earth is rotating about its axis, every object on the surface is also doing the same. As such, all such objects feel the centripetal force radially inward towards the axis of rotation. As such a component of it adds to the already existing gravitational acceleration and changes it. The relation is given as,
$g'=g-{{R}_{e}}{{\omega }^{2}}{{\cos }^{2}}\lambda $
where, g is the true standard value of gravitational acceleration and g’ is its new value. The angular velocity of earth is ω and $\lambda $ is the latitude. According to this, the effective value of gravitational acceleration is minimum at the equator ($\lambda $= 0) and maximum at poles ($\lambda $= 90). Another factor to be considered is the shape of earth. Earth is an ellipsoid. Thus, its radius is more at the equator and less at the poles. Therefore, the value of g is again minimum at equator and maximum at poles.
Therefore the acceleration due to gravity on the surface of earth will be maximum at poles and minimum at equator.
Note:
Students should intuitively remember the various factors on which the acceleration due to gravity depends. They should also keep in mind the direction of other forces acting.
Complete answer:
Newton's law of gravitation states that two objects having masses attract each other by a force proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
where G is the universal gravitation constant. Considering the mass of earth to be $M_e$, and its radius be $R_e$, the acceleration due to gravity g at the surface will be given by,
$g=\dfrac{G{{M}_{e}}}{R_{e}^{2}}$
Now, there are other factors which influence the acceleration due gravity. One of the factors is the latitude of the position. Since the earth is rotating about its axis, every object on the surface is also doing the same. As such, all such objects feel the centripetal force radially inward towards the axis of rotation. As such a component of it adds to the already existing gravitational acceleration and changes it. The relation is given as,
$g'=g-{{R}_{e}}{{\omega }^{2}}{{\cos }^{2}}\lambda $
where, g is the true standard value of gravitational acceleration and g’ is its new value. The angular velocity of earth is ω and $\lambda $ is the latitude. According to this, the effective value of gravitational acceleration is minimum at the equator ($\lambda $= 0) and maximum at poles ($\lambda $= 90). Another factor to be considered is the shape of earth. Earth is an ellipsoid. Thus, its radius is more at the equator and less at the poles. Therefore, the value of g is again minimum at equator and maximum at poles.
Therefore the acceleration due to gravity on the surface of earth will be maximum at poles and minimum at equator.
Note:
Students should intuitively remember the various factors on which the acceleration due to gravity depends. They should also keep in mind the direction of other forces acting.
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