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MCQEasyJEE 2023Variation with Altitude & Depth

JEE Physics 2023 Question with Solution

A body weight WW, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:

  • A

    W91\frac{W}{91}

  • B

    W100\frac{W}{100}

  • C

    W9\frac{W}{9}

  • D

    W3\frac{W}{3}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Weight at earth's surface is WW. The body rises to a height h=9Rh = 9R above the earth, where RR is the earth's radius.

Find: The new weight of the body at that height.

The distance from the center of the earth at the new position is

R+h=R+9R=10RR + h = R + 9R = 10R
A circular earth is shown with a radius marked R from the center to the surface, and a point outside along the same line at height 9R above the surface.

Weight depends on gravitational acceleration, and gravitational acceleration varies inversely as the square of the distance from the earth's center:

g′=g(RR+h)2g' = g\left(\frac{R}{R+h}\right)^2

Substituting h=9Rh = 9R,

g′=g(R10R)2=g100g' = g\left(\frac{R}{10R}\right)^2 = \frac{g}{100}

Since

W=mgW = mg

and

W′=mg′W' = mg'

we get

W′=m(g100)=mg100=W100W' = m\left(\frac{g}{100}\right) = \frac{mg}{100} = \frac{W}{100}

Therefore, the weight of the body at that height is W100\frac{W}{100}. The solution concludes the correct option is A, although this value corresponds to source option B.

Using inverse-square dependence of weight

Given: Initial weight is WW at distance RR from the earth's center. Final height above the surface is 9R9R.

Find: Weight at the new location.

A circular earth is shown with a line from its center to an external point, and the full center-to-point distance is labeled 10R.

At the new point, the body is at distance 10R10R from the earth's center. Hence,

g′=GM(10R)2=1100GMR2=g100g' = \frac{GM}{(10R)^2} = \frac{1}{100}\frac{GM}{R^2} = \frac{g}{100}

Therefore,

W′=mg′=mg100=W100W' = mg' = m\frac{g}{100} = \frac{W}{100}

So the required weight is W100\frac{W}{100}.

Common mistakes

  • Using the height above the surface as the distance from the earth's center is incorrect. The inverse-square law uses distance from the center, so use R+h=10RR+h = 10R, not 9R9R.

  • Assuming weight decreases linearly with height is wrong. Weight depends on 1r2\frac{1}{r^2}, so the square of the distance ratio must be used.

  • Confusing g′g' with W′W' leads to missing the step W′=mg′W' = mg'. First find the new gravitational acceleration, then multiply by mass.

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