An object weighs 200N at the surface of earth. Find the weight at a depth of \(\frac{R}{2}\) , where R is radius of earth
- 100N
- 300N
- 50N
- 150N
The Correct Option is A
Approach Solution - 1
Where, m is the mass of the object and g is acceleration due to gravity on the surface of the earth.
The variation of acceleration due to gravity (g) with the depth (d) from the surface of the Earth is given by
gd=g1-dR
Where
gd is the acceleration due to gravity at depth d
R is the radius of the Earth
At depth d = \(\frac{R}{2}\), acceleration due to gravity is given by
gd=g1-\(\frac{R}{2R}\)=g2
Multiplying both sides by mass of the object, then we get
mgd = \(\frac{mg}{2}\)
But, mg = 200 N
Therefore, Weight of the object at depth R/2 is given by
mgd = \(\frac{200}{2}\) = 100 N
Answer. A
Approach Solution -2
Acceleration due to gravity at depth d from surface of earth
Multiplying by mass 'm' on both sides,
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Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
Newton’s Law of Gravitation:
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
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Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]