If the distance between the earth and the sun were half its present value, the number of days in a y

Q: Why does the year get shorter when the distance decreases? Shouldn't it just be a smaller circle at the same speed?

No! As the Earth gets closer to the Sun, the gravitational force of attraction increases significantly because gravity follows an inverse-square law ($F \propto 1/r^2$). To maintain a stable circular orbit and not spiral into the Sun, the Earth must move much faster. A faster speed combined with a shorter path length results in a much shorter year!

Q: Can we use Newton's Law of Gravitation to derive Kepler's Third Law?

Absolutely! For a circular orbit, the gravitational force provides the necessary centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$. Since orbital speed is $v = \frac{2\pi r}{T}$, substituting this gives $\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}$, which simplifies to $T^2 = \left(\frac{4\pi^2}{GM}\right) r^3$. This is exactly Kepler's Third Law!

Q: What would happen to the seasons if the year became 129 days?

The seasons would still occur because they are caused by the tilt of the Earth's axis (which is $23.5^\circ$), not the distance from the Sun. However, each season would be much shorter—lasting only about 32 days instead of the usual 90 days!

Q: Why did we use 365 days instead of 365.25 days?

Using 365 days is a standard approximation in physics problems unless high precision is specified. Even if you use 365.25 days, the calculation yields $365.25 / 2.828 \approx 129.1$ days, which still rounds beautifully to the nearest integer option of 129 days.

Q: Does the mass of the Earth affect the length of the year?

Surprisingly, no! In our derivation, the mass of the planet $m$ cancels out on both sides of the force equation. The orbital period depends only on the mass of the central body (the Sun) and the orbital radius. This is why a tiny satellite and a massive planet at the same distance would have the exact same orbital period!

Question

If the distance between the earth and the sun were half its present value, the number of days in a year would have been

Options:

(A) 64.5

(B) 129

(C) 182.5

(D) 730

Sigma Percentile · Accepted Answer

Imagine standing on Earth, watching the seasons change over the course of 365 days. This familiar rhythm is dictated by our distance from the Sun. But what if we were to suddenly alter this cosmic arrangement? What if the Earth were pulled closer, halving its distance to the Sun? How would our calendar respond to this dramatic shift?

This is not just a science fiction scenario; it is a classic physics problem that invites us to explore the fundamental laws governing planetary motion. Let us embark on a journey to solve this mystery using the elegant principles of gravitation.

### The Cosmic Scale and Kepler's Insight

In the early 17th century, Johannes Kepler analyzed decades of precise astronomical observations made by Tycho Brahe. From this data, he formulated three laws that beautifully describe how planets orbit the Sun. To solve our problem, we must turn to **Kepler's Third Law**, also known as the **Law of Periods**.

Kepler discovered a profound mathematical harmony in the solar system: the square of a planet's orbital period is directly proportional to the cube of its average distance from the Sun.

$$T^2 \propto r^3$$

Here, $T$ represents the time period of revolution (the length of a year), and $r$ represents the radius of the circular orbit (or the semi-major axis of an elliptical orbit).

### Setting Up the Ratio

To see how a change in distance affects the time period, we can write Kepler's Third Law as a ratio for two different states. Let the original state of the Earth be represented by subscript 1, and the new state by subscript 2.

Taking the square root on both sides of the proportionality gives us:

$$T \propto r^{3/2}$$

Now, we can write the ratio of the new time period $T_2$ to the original time period $T_1$:

$$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{3/2}$$

This simple yet powerful equation is our gateway to the solution. It tells us that the change in the length of a year is not linear; it depends on the fractional power of the change in distance.

### Substituting the Cosmic Parameters

Let us look at the values given in our problem. We are told that the new distance is half of the present value:

$$r_2 = \frac{r_1}{2} \implies \frac{r_2}{r_1} = \frac{1}{2}$$

We also know that our current year, $T_1$, is approximately 365 days. Let us substitute these values into our ratio equation:

$$T_2 = T_1 \left(\frac{1}{2}ight)^{3/2}$$

$$T_2 = 365 	imes \left(\frac{1}{2}ight)^{3/2}$$

### The Mathematical Computation

Now, we must carefully evaluate the term $\left(\frac{1}{2}ight)^{3/2}$. This fractional exponent can be broken down into a cube and a square root:

$$\left(\frac{1}{2}ight)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{\sqrt{2^3}} = \frac{1}{\sqrt{8}}$$

We can simplify the square root of 8 by factoring out the perfect square 4:

$$\sqrt{8} = \sqrt{4 	imes 2} = 2\sqrt{2}$$

So, our expression becomes:

$$T_2 = \frac{365}{2\sqrt{2}}$$

Let us use the standard approximation for the square root of 2, which is $\sqrt{2} \approx 1.414$:

$$2\sqrt{2} \approx 2 	imes 1.414 = 2.828$$

Now, we substitute this back to find the new time period:

$$T_2 \approx \frac{365}{2.828} \approx 129.05	ext{ days}$$

### Celebrating the Result

Our calculation reveals that if the Earth-Sun distance were halved, a year would last only about **129 days**!

This is a dramatic reduction from our current 365-day year. It means that the Earth would have to complete its orbit much faster to balance the stronger gravitational pull of the Sun at this closer distance.

Comparing our result with the given options, we find that it matches perfectly with option (b).

Therefore, the correct option is **(b)**.

Animated Solution for Physics - Gravitation: If the distance between the earth and the sun were half its present value, the number of days in a year would have been

Select Answer:

Visualized Solution

Visualizing the Planetary Orbits

Kepler's Third Law of Planetary Motion

Expressing Time Period in terms of Radius

Substituting the Given Values

Evaluating the Fractional Power

Numerical Approximation

Calculating the New Number of Days

Exploring Further: Orbital Velocity

The Sigma Insight: Kepler's Laws of Planetary Motion

The Cosmic Scale and Kepler's Insight

Setting Up the Ratio

Substituting the Cosmic Parameters

The Mathematical Computation

Celebrating the Result

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