Why the weight of the body is less inside the earth than on the surface?
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It is easier to understand this phenomenon using study of gravity inside a hollow shell. The effective gravity inside any hollow shell (from the material of the shell) is zero.
For any point inside the surface of the earth at radius ‘r’, there are two parts of earth acting on the point differently:
1. All the mass of the earth above the radius ‘r’. This portion forms a hollow shell, whose effective gravity at any point inside it will be zero.
2. The inner sphere of radius ‘r’. As we reduce the size of a uniform sphere (hence the distance from center), the gravitational force increases quadratically, but the mass reduces by the power of 3 because `m is proportional to r^3 while F is proportional to 1/r^2′. Thus the gravitational force reduces as we go deeper into the earth, and eventually reducing the weight of the object.
For any point inside the surface of the earth at radius ‘r’, there are two parts of earth acting on the point differently:
1. All the mass of the earth above the radius ‘r’. This portion forms a hollow shell, whose effective gravity at any point inside it will be zero.
2. The inner sphere of radius ‘r’. As we reduce the size of a uniform sphere (hence the distance from center), the gravitational force increases quadratically, but the mass reduces by the power of 3 because `m is proportional to r^3 while F is proportional to 1/r^2′. Thus the gravitational force reduces as we go deeper into the earth, and eventually reducing the weight of the object.
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Explanation:
W=mg
and g decreases when we go inside the earth
and W is directly proportional to g
so weight is less inside the earth than on the surface
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