Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period

Q: Why do we still use $mv^2/R$ for centripetal force if the gravity law changed?

The formula $mv^2/R$ is a purely kinematic requirement for *any* object moving in a circle, regardless of what force is pushing or pulling it. The change in gravity only changes *what* provides that force, not the requirement itself.

Q: What happens to the mass of the planet $m$?

It cancels out from both sides of the equation $mv^2/R = GMm/R^n$. This means the orbital speed and time period depend only on the mass of the central star and the radius of the orbit, not on how heavy the planet is.

Q: How did $R$ become $R^2$ when moved inside the square root?

This is a basic rule of algebra. Since $\sqrt{x^2} = x$ (for positive $x$), you can rewrite $R$ as $\sqrt{R^2}$. When you multiply $\sqrt{R^2}$ by $\sqrt{R^{n-1}/GM}$, you can combine them under a single square root.

Q: Why is $2\pi/\sqrt{GM}$ considered a constant?

$G$ is the universal gravitational constant, and $M$ is the mass of the central star (like our Sun). Unless the star is losing or gaining massive amounts of mass, this entire term remains constant for any planet orbiting it.

Q: Does this mean Kepler's laws are wrong?

No! Kepler's laws are specifically derived for our universe where gravity follows an inverse square law ($n=2$). This question is a hypothetical 'what if' scenario to test your ability to derive relationships from scratch.

Question

Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

Options:

(A) $R^{\left(\frac{n+1}{2}\right)}$

(B) $R^{\left(\frac{n-1}{2}\right)}$

(C) $R^n$

(D) $R^{\left(\frac{n-2}{2}\right)}$

Sigma Percentile · Accepted Answer

### The Hypothetical Universe

Imagine a universe where the laws of physics are slightly tweaked. In our real universe, Newton's Law of Universal Gravitation states that the force between two masses is inversely proportional to the square of the distance between them ($F \propto 1/R^2$). But what if gravity was a bit more exotic? What if the force varied inversely as the $n^{	ext{th}}$ power of the distance?

This is a classic thought experiment in physics that tests your fundamental understanding of orbital mechanics. Let's dive into how this hypothetical change affects the time period of a planet revolving around a star.

### Setting Up the Dynamics

Consider a planet of mass $m$ in a circular orbit of radius $R$ around a star of mass $M$. The modified gravitational force acting on the planet is given by:

$$F_g = \frac{GMm}{R^n}$$

For the planet to maintain its circular orbit, it requires a centripetal force. This force is entirely provided by the gravitational pull of the star. The centripetal force required for an object moving with speed $v$ in a circle of radius $R$ is:

$$F_c = \frac{mv^2}{R}$$

By equating the gravitational force to the centripetal force, we establish the dynamic equilibrium of the orbit:

$$\frac{mv^2}{R} = \frac{GMm}{R^n}$$

### Finding the Orbital Velocity

Notice how the mass of the planet, $m$, appears on both sides of the equation. This is a beautiful feature of gravity—the orbital kinematics are independent of the orbiting body's mass! Canceling $m$ and rearranging the equation to solve for $v^2$, we get:

$$v^2 = \frac{GM}{R^{n-1}}$$

Taking the square root gives us the orbital velocity:

$$v = \sqrt{\frac{GM}{R^{n-1}}}$$

### Deriving the Time Period

The time period $T$ is the time it takes for the planet to complete one full revolution. It is simply the circumference of the orbit divided by the orbital speed:

$$T = \frac{2\pi R}{v}$$

Now, we substitute our expression for $v$ into this formula:

$$T = \frac{2\pi R}{\sqrt{\frac{GM}{R^{n-1}}}}$$

When we bring the denominator's fraction up, it flips:

$$T = 2\pi R \sqrt{\frac{R^{n-1}}{GM}}$$

To simplify this, we can bring the $R$ that is outside the square root inside. When $R$ goes inside a square root, it becomes $R^2$.

$$T = 2\pi \sqrt{\frac{R^2 \cdot R^{n-1}}{GM}}$$

Using the laws of exponents, we add the powers of $R$: $2 + (n - 1) = n + 1$. This yields:

$$T = 2\pi \sqrt{\frac{R^{n+1}}{GM}}$$

### The Final Proportionality

We can rewrite the square root as a fractional exponent:

$$T = \frac{2\pi}{\sqrt{GM}} R^{\frac{n+1}{2}}$$

Since $2\pi$ and $\sqrt{GM}$ are constants for a given star system, we can conclude that the time period is proportional to $R$ raised to the power of $(n+1)/2$:

$$T \propto R^{\frac{n+1}{2}}$$

**Sanity Check:** Let's test our formula with the real universe where $n = 2$. Plugging $n = 2$ into our result gives $T \propto R^{(2+1)/2} = R^{3/2}$. Squaring both sides gives $T^2 \propto R^3$, which is exactly **Kepler's Third Law of Planetary Motion**! It is always deeply satisfying when a generalized mathematical derivation perfectly collapses into a known physical law.

Animated Solution for Physics - Gravitation: Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

Select Answer:

Visualized Solution

\text{Orbital Setup}

\text{Modified Gravitational Force}

\text{Centripetal Force}

\text{Equating Forces}

\text{Orbital Velocity}

\text{Time Period Formula}

\text{Substituting } v

\text{Simplifying the Expression}

\text{Proportionality}

\text{Kepler's Third Law Check}

The Sigma Insight: Kepler's Laws of Planetary Motion

The Hypothetical Universe

Setting Up the Dynamics

Finding the Orbital Velocity

Deriving the Time Period

The Final Proportionality

Similar Questions

Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$ . If the gravitational force of attraction between the planet and the star is proportional to $R^{- 5/2}$ , then

The time period of a satellite of earth is $5 h$ . If the separation between the earth and the satellite is increased to $4$ times the previous value, the new time period will become

A geostationary satellite is orbiting around an arbitrary planet $P$ at a height of $11 R$ above the surface of $P$ , $R$ being the radius of $P$ . The time period of another satellite in hours at a height of $2 R$ from the surface of $P$ is ............ . $P$ has the time period of 24 h.

The time period of a satellite in a circular orbit of radius $R$ is $T$ . The period of another satellite in a circular orbit of radius $9 R$ is

A satellite is launched into a circular orbit of radius $R$ around Earth, while a second satellite is launched into a circular orbit of radius $1.02 R$ . The percentage difference in the time periods of the two satellites is

If the angular momentum of a planet of mass $m$ , moving around sun in a circular orbit is $L$ about the centre of the sun, its areal velocity is

The angular momentum of a planet of mass $M$ moving around the Sun in an elliptical orbit is $L$ . The magnitude of the areal velocity of the planet is

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