The value of the acceleration due to gravity is g1 at a height h = R/2 (where, R = radius of the ear

Q: Why can't I just use the formula $g_h = g(1 - 2h/R)$? It's much faster!

That formula is a binomial approximation. It is only valid when $h$ is very, very small compared to $R$ (typically $h < 5\% \text{ of } R$). Here, $h = 50\% \text{ of } R$, so the approximation introduces a massive error. Always use the exact formula $g_h = \frac{GM}{(R+h)^2}$ for large heights.

Q: Why does gravity decrease when we go deep into the Earth? Shouldn't we be getting closer to the center?

While you are getting closer to the center, the amount of Earth's mass 'below' you that is pulling you inward is also decreasing. The mass 'above' you exerts forces that cancel out. The loss of effective mass outweighs the decrease in distance, causing a net linear decrease in gravity.

Q: Is the formula for depth $g_d = g(1 - d/R)$ an approximation too?

No, it is an exact formula, provided we assume the Earth is a perfect sphere with uniform density. Unlike the height formula, there is no binomial expansion involved in its derivation.

Q: What happens to the $GM$ terms in the equation?

Since $GM$ represents the Universal Gravitational Constant multiplied by the mass of the Earth, it is a non-zero constant. Because it appears as a multiplier on both sides of our equated expressions, we can safely divide both sides by $GM$ to cancel it out.

Q: How do I know when to use the exact formula versus the approximation in exams?

As a rule of thumb, if the height $h$ is given in terms of $R$ (like $R/2$, $R/4$, $2R$), always use the exact formula. If $h$ is given as a small numerical value like $10 \text{ km}$ or $32 \text{ km}$ (where $R = 6400 \text{ km}$), you can safely use the approximation to save time.

Question

The value of the acceleration due to gravity is $g_1$ at a height $h = \frac{R}{2}$ (where, $R = $ radius of the earth) from the surface of the earth. It is again equal to $g_1$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

Options:

(A) $\frac{7}{9}$

(B) $\frac{1}{3}$

(C) $\frac{4}{9}$

(D) $\frac{5}{9}$

Sigma Percentile · Accepted Answer

### The Trap of Approximations

When dealing with the variation of acceleration due to gravity ($g$) with height, students often reflexively reach for the formula $g_h = g\left(1 - \frac{2h}{R}\right)$. However, this is a classic trap! This formula is a binomial approximation that is strictly valid only when the height $h$ is extremely small compared to the radius of the Earth $R$ (i.e., $h \ll R$).

In this problem, we are given $h = \frac{R}{2}$. This is a massive height—half the radius of the Earth! Using the approximation here would lead to a disastrously wrong answer. We must rely on the fundamental, exact formula.

### Setting Up the Master Equations

The exact formula for acceleration due to gravity at a height $h$ above the Earth's surface is derived directly from Newton's Law of Universal Gravitation:

$$g_h = \frac{GM}{(R+h)^2}$$

On the other hand, the variation of gravity with depth $d$ is linear, assuming the Earth is a sphere of uniform density. The formula for gravity at a depth $d$ is:

$$g_d = \frac{GM}{R^3}(R-d) = \frac{GM}{R^2}\left(1 - \frac{d}{R}\right)$$

### Equating and Solving

The problem states that the gravity at height $h$ is exactly equal to the gravity at depth $d$. Let's set our two master equations equal to each other and substitute $h = \frac{R}{2}$:

$$\frac{GM}{\left(R + \frac{R}{2}\right)^2} = \frac{GM}{R^2}\left(1 - \frac{d}{R}\right)$$

First, let's simplify the denominator on the left-hand side. Adding $R$ and $\frac{R}{2}$ gives us $\frac{3R}{2}$. Squaring this term yields $\frac{9R^2}{4}$. Substituting this back into our equation:

$$\frac{GM}{\frac{9R^2}{4}} = \frac{GM}{R^2}\left(1 - \frac{d}{R}\right)$$

$$\frac{4GM}{9R^2} = \frac{GM}{R^2}\left(1 - \frac{d}{R}\right)$$

### The Final Calculation

Notice the beautiful symmetry here. The term $\frac{GM}{R^2}$ (which is simply $g$ at the surface) appears on both sides of the equation. We can cleanly cancel it out, leaving us with a pure, dimensionless algebraic relation:

$$\frac{4}{9} = 1 - \frac{d}{R}$$

Now, it's just a matter of simple rearrangement to isolate the ratio we are looking for:

$$\frac{d}{R} = 1 - \frac{4}{9}$$

**$$\frac{d}{R} = \frac{5}{9}$$**

And there we have it! By carefully choosing the exact formula instead of the approximation, the math flows smoothly to the correct answer.

Select Answer:

Visualized Solution

$g_{h} = g_{d} = g_{1}$

$g_{h} = \frac{GM}{( R + h ) ^{2}}$

$g_{d} = \frac{GM}{R ^{2}} (1 - \frac{d}{R})$

$\frac{GM}{( R + \frac{R}{2} ) ^{2}} = \frac{GM}{R ^{2}} (1 - \frac{d}{R})$

$\frac{GM}{( \frac{3 R}{2} ) ^{2}} = \frac{GM}{R ^{2}} (1 - \frac{d}{R})$

$\frac{4 GM}{9 R ^{2}} = \frac{GM}{R ^{2}} (1 - \frac{d}{R})$

$\frac{4}{9} = 1 - \frac{d}{R}$

$\frac{d}{R} = \frac{5}{9}$

The Sigma Insight: Acceleration due to Gravity and its Variation

The Trap of Approximations

Setting Up the Master Equations

Equating and Solving

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } gh​=gd​=g1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } gh​=(R+h)2GM​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } gd​=R2GM​(1−Rd​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (R+2R​)2GM​=R2GM​(1−Rd​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (23R​)2GM​=R2GM​(1−Rd​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9R24GM​=R2GM​(1−Rd​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 94​=1−Rd​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rd​=95​