The formula is a first-order binomial approximation derived assuming (typically ). Here , which is far too large for the approximation to be valid.

JEE MainPhysics

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Gravitation: JEE Main Physics Question with Solution

The weight of a body on the surface of the earth is

100\text{ N}

. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

100\text{ N}

64\text{ N}

50\text{ N}

25\text{ N}

Step-by-step solutionView

Correct answer

At a height $h = R/4$ , the distance from Earth's center becomes $5R/4$ , reducing the gravitational force to $(4/5)^2 \times 100\text{ N} = 64\text{ N}$ .

Option analysis

Why each option works or fails

A ·

100\text{ N}

Assuming that gravitational force remains constant near Earth's surface regardless of significant altitude. Recognize that a height of $R/4$ is comparable to Earth's radius, so the inverse-square law $F \propto 1/r^2$ causes a noticeable reduction.

B ·

64\text{ N}

None. This is the correct calculation using the inverse-square law with orbital distance $r = R + R/4 = 5R/4$ . Correctly compute $F = F_0 / (1 + h/R)^2 = 100 / (1 + 1/4)^2 = 100 / (25/16) = 64\text{ N}$ .

C ·

50\text{ N}

Using an incorrect linear approximation or miscalculating the squared denominator as approximately 2. Apply the exact inverse-square relation $F \propto 1/r^2$ without approximating, since $h$ is not negligible compared to $R$ .

D ·

25\text{ N}

Taking the distance from Earth's center to be $h = R/4$ directly, rather than $r = R + h = 5R/4$ , or applying an incorrect power relation. Measure distance $r$ from the center of mass of Earth ( $r = R + h$ ), not from the surface.

Reviewed route

Solution

StepWorking

01given

Weight on surface $W = mg = 100\text{ N}$ , height above surface $h = R/4$ , where $R$ is the radius of the Earth.

02find

Gravitational force (weight) $W' = mg'$ at height $h$ .

03visualise

A body of mass $m$ is initially on the Earth's surface at distance $R$ from the center, and is moved to a distance $r = R + h = R + R/4 = 5R/4$ from the center.

04strategise

Use the inverse-square law for gravity outside the Earth: $g' = g \left(\frac{R}{R+h}\right)^2$ . Therefore, $W' = W \left(\frac{R}{R+h}\right)^2$ .

05execute

Substitute $h = R/4$ and $W = 100\text{ N}$ : $W' = 100 \times \left(\frac{R}{R + R/4}\right)^2 = 100 \times \left(\frac{4}{5}\right)^2 = 100 \times \frac{16}{25} = 64\text{ N}$ .

✓verify

Since $h > 0$ , the weight must decrease ( $< 100\text{ N}$ ). At $h = R/4$ , $r = 1.25R$ , so $W' = 100 / (1.25)^2 = 100 / 1.5625 = 64\text{ N}$ . Units and magnitude are consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can't we use

g' = g\left(1 - \frac{2h}{R}\right)

The formula $g' = g(1 - 2h/R)$ is a first-order binomial approximation derived assuming $h \ll R$ (typically $h < 5\% R$ ). Here $h = R/4 = 25\% R$ , which is far too large for the approximation to be valid.

Answer

At a height $h = R/4$ , the distance from Earth's center becomes $5R/4$ , reducing the gravitational force to $(4/5)^2 \times 100\text{ N} = 64\text{ N}$ .

Why each option works or fails

A: $100\text{ N}$ - Assuming that gravitational force remains constant near Earth's surface regardless of significant altitude. Recognize that a height of $R/4$ is comparable to Earth's radius, so the inverse-square law $F \propto 1/r^2$ causes a noticeable reduction.
B · correct: $64\text{ N}$ - None. This is the correct calculation using the inverse-square law with orbital distance $r = R + R/4 = 5R/4$ . Correctly compute $F = F_0 / (1 + h/R)^2 = 100 / (1 + 1/4)^2 = 100 / (25/16) = 64\text{ N}$ .
C: $50\text{ N}$ - Using an incorrect linear approximation or miscalculating the squared denominator as approximately 2. Apply the exact inverse-square relation $F \propto 1/r^2$ without approximating, since $h$ is not negligible compared to $R$ .
D: $25\text{ N}$ - Taking the distance from Earth's center to be $h = R/4$ directly, rather than $r = R + h = 5R/4$ , or applying an incorrect power relation. Measure distance $r$ from the center of mass of Earth ( $r = R + h$ ), not from the surface.

Step-by-step solution

given: Weight on surface $W = mg = 100\text{ N}$ , height above surface $h = R/4$ , where $R$ is the radius of the Earth.
find: Gravitational force (weight) $W' = mg'$ at height $h$ .
visualise: A body of mass $m$ is initially on the Earth's surface at distance $R$ from the center, and is moved to a distance $r = R + h = R + R/4 = 5R/4$ from the center.
strategise: Use the inverse-square law for gravity outside the Earth: $g' = g \left(\frac{R}{R+h}\right)^2$ . Therefore, $W' = W \left(\frac{R}{R+h}\right)^2$ .
execute: Substitute $h = R/4$ and $W = 100\text{ N}$ : $W' = 100 \times \left(\frac{R}{R + R/4}\right)^2 = 100 \times \left(\frac{4}{5}\right)^2 = 100 \times \frac{16}{25} = 64\text{ N}$ .
verify: Since $h > 0$ , the weight must decrease ( $< 100\text{ N}$ ). At $h = R/4$ , $r = 1.25R$ , so $W' = 100 / (1.25)^2 = 100 / 1.5625 = 64\text{ N}$ . Units and magnitude are consistent.

Shortcut:

When to use it: When distance ratio from the center is straightforward.

given: $W_1 = 100\text{ N}$ at $r_1 = R$ , target distance $r_2 = R + R/4 = 1.25 R = \frac{5}{4}R$ .

find: Weight $W_2$ at $r_2$ .

visualise: Gravitational force scales as $F \propto 1/r^2$ .

strategise: $W_2 = W_1 \left(\frac{r_1}{r_2}\right)^2$ .

execute: $W_2 = 100 \times \left(\frac{4}{5}\right)^2 = 100 \times \frac{16}{25} = 64\text{ N}$ .

verify: $64\text{ N} < 100\text{ N}$ , matches option (1).

Gravitation: JEE Main Physics Question with Solution

What feels right?

At a height $h = R/4$ , the distance from Earth's center becomes $5R/4$ , reducing the gravitational force to $(4/5)^2 \times 100\text{ N} = 64\text{ N}$ .

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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