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A box weighs 432 N on the surface of Earth. If R is the radius of the Earth, what is the weight of the box at a height of 3R above the Earth's surface?
1. 432 N
2. 0 N
3. 48 N
4. 27 N

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Best answer
Correct Answer - Option 4 : 27 N

The correct answer is option 4) i.e. 27 N

CONCEPT:

  • Acceleration due to gravity on the surface of Earth of mass M and radius Re is denoted by g.
    • It has an approximated uniform value of 9.8 m/s2 on the surface of Earth.

The acceleration due to gravity at a depth 'd' below the surface of Earth is given by 

\(\Rightarrow g' = g(1- \frac{d}{R_e})\)

  • The acceleration due to gravity at a height 'h' above the surface of Earth is given by 

\(\Rightarrow g'' = g(1+ \frac{h}{R_e})^{-2}\)

  • Weight: Weight is defined as the force with which an object is pulled towards the Earth due to gravity.

It is given by:

Weight, W = mg

Where m is the mass of the object and g is the acceleration due to gravity.

CALCULATION:

Given that: 

Weight of the object on Earth, W = 432 N = mg      ----(1)

Acceleration due to gravity at height h = 3R from the surface of Earth, \(g'' = g(1+ \frac{h}{R_e})^{-2}\)

\(\Rightarrow g'' = g(1+ \frac{3R}{R})^{-2} = g (4)^{-2} = \frac{g}{16} \: m/s^2\)

Using, W = mg

The weight of the box at 3R, \(W'' = mg''= m\times \frac{g}{16} = \frac{W}{16} \)      ----(2)

Substituting (1) in (2),

\(W'' =\frac{423}{16} = 27\: N\)

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