Q.In the Rutherford's nuclear model of the atom, the nucleus (radius about $10^{-15}\ \text{m}$) is analogous to the sun about which the electron moves in orbit (radius $\approx 10^{-10}\ \text{m}$) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is? The radius of earth's orbit is about $1.5 \times 10^{11}\ \text{m}$. The radius of sun is taken as $7 \times 10^{8}\ \text{m}$.

Compare the two "orbit-radius to central-body-radius" ratios. For the atom, $r_{orbit}/r_{nucleus}=10^{-10}/10^{-15}=10^{5}$. For the real solar system, $R_{orbit}/R_{sun}=1.5\times10^{11}/7\times10^{8}\approx214$. To make the solar system atom-like (same $10^5$ ratio) with the Sun's radius fixed, the orbit would need $R'=10^{5}\times7\times10^{8}=7\times10^{13}\ \text{m}$, about $470$ times the real $1.5\times10^{11}\ \text{m}$.

Concept understanding

The scale of an orbiting system is captured by the ratio of the orbit radius to the radius of the central body. We compare that ratio for the atom with the same ratio for the solar system.

Working it out

1. The atom's ratio.

The electron orbits at $10^{5}$ times the nuclear radius.

2. The real solar system's ratio.

3. Rescale to be atom-like. Keeping the Sun (the nucleus analogue) fixed, the orbit must satisfy $R'_{orbit}/R_{sun}=10^{5}$:

4. Compare with the true orbit.

Because the nucleus is tinier relative to the electron's orbit ($10^5$) than the Sun is relative to Earth's orbit ($214$), reproducing the atomic proportion forces the orbit to swell by a factor of about $470$.

Method: Direct Proportion (Scale Factor Comparison)

The core idea is that the atom and the solar system are being treated as scale models of each other. In the atom, the nucleus is the “sun” and the electron’s orbit is the “earth’s orbit.” We find the ratio of orbit radius to nucleus radius in the atom, then apply that same ratio to the sun to see what the earth’s orbit would be if the solar system were scaled the same way.

Steps

1. Find the scale factor in the atom. The electron orbits at radius $10^{-10}\ \text{m}$ around a nucleus of radius $10^{-15}\ \text{m}$. The ratio is:

So the orbit is $10^{5}$ times larger than the nucleus.

2. Apply the same scale factor to the solar system. The sun’s radius is given as $7 \times 10^{8}\ \text{m}$. If the earth’s orbit were proportioned like the atom, its radius would be:

3. Compare with the actual earth–sun distance. The actual radius of earth’s orbit is $1.5 \times 10^{11}\ \text{m}$. Since $7 \times 10^{13}\ \text{m}$ is much larger than $1.5 \times 10^{11}\ \text{m}$, the scaled orbit is farther away.

Final answer:

The earth would be farther away from the sun than it actually is.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing which ratio to compare

Students often compare the nucleus radius to the electron orbit radius directly, or compare the sun's radius to the earth's orbit radius — without realising the analogy works by matching the same kind of ratio on both sides.

How to avoid: The analogy says: nucleus : electron orbit is like sun : earth's orbit. So you must set up:

You are not comparing nucleus to sun directly. You are comparing the proportions of the two systems.

Mistake 2: Forgetting to solve for the new earth-sun distance

Many students compute the ratio $10^{-5}$ and then stop, or they multiply the wrong quantities. They might calculate $7 \times 10^8 \times 10^{-5}$ instead of dividing.

How to avoid: Once you set up the proportion:

Cross-multiply carefully:

Mistake 3: Misinterpreting "closer or farther"

After finding $R_{\text{new}} = 7 \times 10^{13}\ \text{m}$, students compare it to the sun's radius instead of the actual earth-sun distance ($1.5 \times 10^{11}\ \text{m}$).

How to avoid: The question asks: compared to the actual earth-sun distance, is the new distance larger or smaller?

Actual: $1.5 \times 10^{11}\ \text{m}$

New: $7 \times 10^{13}\ \text{m}$

Since $7 \times 10^{13} > 1.5 \times 10^{11}$, the earth would be farther away.

Mistake 4: Using the wrong units or forgetting powers of ten

Students sometimes treat $10^{-15}$ and $10^{-10}$ as if they were $10^{-15}\ \text{m}$ and $10^{-10}\ \text{m}$ but then drop the exponents when dividing, or misplace decimal points with $7 \times 10^8$.

How to avoid: Write every number in scientific notation before calculating. Do not approximate $10^{-15}/10^{-10}$ as $10^{-5}$ in your head without writing it down — one slip and the answer is off by a factor of 10.

Mistake 5: Thinking the analogy means "same size" not "same proportion"

Some students assume the nucleus is the sun and the electron is the earth, so they directly compare $10^{-15}$ to $7 \times 10^8$ and conclude the atom is much smaller — missing the point entirely.

How to avoid: The word "analogous" means the relationship is similar, not the sizes. The atom's nucleus is tiny compared to its orbit; the question asks: if the solar system had that same ratio, what would happen?

The scaled solar system has a much larger orbit because the sun stays the same size but the ratio must shrink to match the atom's extreme proportion.

Final answer: The earth would be farther away from the sun than it actually is — at a distance of $7 \times 10^{13}\ \text{m}$ instead of $1.5 \times 10^{11}\ \text{m}$.