Animated Solution for Physics - Gravitation: The maximum and minimum distance of a comet from the Sun are $1.6\times 10^{12}$ m and $8.0\times 10^{10}$ m, respectively. If the speed of the comet at the nearest point is $6\times 10^4{\text{ ms}}^{-1}$, then the speed at the farthest point is

Comets and Elliptical Orbits

Imagine a comet sweeping through the vast expanse of our solar system. Unlike the nearly circular orbits of planets, comets often travel in highly elongated, elliptical paths. The Sun doesn't sit at the center of this ellipse; instead, it occupies one of the focal points. This geometric reality means that the comet's distance from the Sun changes dramatically over time.

There is a point of closest approach, known as perihelion, where the comet feels the strongest gravitational pull and whips around the Sun at breakneck speed. Conversely, there is a point of maximum distance, called aphelion, where the comet lingers, moving sluggishly before beginning its long fall back inward. Our goal in this problem is to find exactly how slow the comet gets at this farthest point.

Conservation of Angular Momentum

To solve this, we need a powerful physical principle. As the comet travels, the only significant force acting on it is the gravitational pull of the Sun. Because this force always points directly toward the Sun (a central force), it cannot exert any twisting force, or torque, on the comet.

In physics, when the net external torque on a system is zero, the angular momentum of the system remains perfectly conserved. The angular momentum $L$ of a particle of mass $m$ moving with velocity $v$ at a distance $r$ from the origin is given by:

Here, $\theta$ is the angle between the position vector and the velocity vector. At the extreme points of the orbit—the absolute nearest and farthest points—the comet is momentarily moving purely tangentially. This means the velocity vector is exactly perpendicular to the position vector, making $\theta =90^{\circ }$ and $\sin (90^{\circ })=1$.

Because angular momentum is conserved, its value at the farthest point must equal its value at the nearest point. Let's denote the farthest point with subscript 1 and the nearest point with subscript 2. We can write our master equation as:

Notice something beautiful here? The mass of the comet, $m$, appears on both sides of the equation. We can completely cancel it out! This tells us that the orbital kinematics are independent of the comet's mass. We are left with a simple, elegant relationship:

We want to find the speed at the farthest point, $v_1$. Rearranging the equation gives us our raw setup:

Now, it's time to substitute the given values into our equation. We know the maximum distance $r_1=1.6\times 10^{12}$ m, the minimum distance $r_2=8.0\times 10^{10}$ m, and the maximum speed $v_2=6\times 10^4{\text{ ms}}^{-1}$.

When dealing with scientific notation, it's a great habit to group the coefficients and the powers of ten separately to avoid silly mistakes. Let's multiply the coefficients in the numerator first: $6\times 8.0=48$. Next, we combine the powers of ten by adding their exponents: $10^4\times 10^{10}=10^{14}$.

Now, we perform the final division. Dividing the coefficients gives $48/1.6=30$. Subtracting the exponents for the powers of ten gives $10^{14}/10^{12}=10^2$.

Kepler's Legacy

To express our final answer in standard scientific notation, we adjust the decimal point:

This result is not just a number; it's a profound demonstration of Kepler's Second Law of Planetary Motion, which states that a line segment joining a planet (or comet) and the Sun sweeps out equal areas during equal intervals of time. Because the areal velocity $\frac{dA}{dt}=\frac{L}{2m}$ is constant, a larger distance $r$ demands a smaller velocity $v$ to keep the product $vr$ constant. The comet must slow down as it climbs away from the Sun's gravity well, perfectly obeying the laws of cosmic mechanics!