Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 10^{8}\text{ m}$. Show that the mass of Jupiter is about one-thousandth that of the sun.

> [!TLDR] > By applying Kepler's Third Law, which relates the orbital period and radius of a satellite to the mass of its central body, we calculate Jupiter's mass using Io's orbital data. Comparing this to the known mass of the Sun reveals that Jupiter's mass is approximately one-thousandth that of the Sun. To determine the mass of Jupiter and compare it to the Sun's mass, we use Kepler's Third Law of planetary motion. This law, derived from Newton's Law of Universal Gravitation and the concept of centripetal force, provides a direct relationship between the orbital period ($T$) and orbital radius ($r$) of a satellite and the mass ($M$) of the central body it orbits. The fundamental idea is that the gravitational force exerted by Jupiter on Io provides the necessary centripetal force to keep Io in its orbit. By equating these two forces, we can solve for the mass of Jupiter. > [!FORMULA] > The mass $M$ of the central body is given by: > $$M = \frac{4 \pi^2 r^3}{G T^2}$$ > where $r$ is the orbital radius, $T$ is the orbital period, and $G$ is the universal gravitational constant. > [!PROOF] Derivation of Kepler's Third Law > Consider a satellite of mass $m$ orbiting a central body of mass $M$ in a circular path of radius $r$. > > 1. **Gravitational Force:** The force of gravity attracting the satellite to the central body is given by Newton's Law of Universal Gravitation: > $$F_g = \frac{G M m}{r^2}$$ > > 2. **Centripetal Force:** For the satellite to maintain a circular orbit, it must experience a centripetal force directed towards the center of the orbit. If the satellite's orbital speed is $v$, this force is: > $$F_c = \frac{m v^2}{r}$$ > > 3. **Equating Forces:** In a stable orbit, the gravitational force provides the centripetal force: > $$\frac{G M m}{r^2} = \frac{m v^2}{r}$$ > > 4. **Relating Speed to Period:** For a circular orbit, the speed $v$ is the distance traveled (circumference $2 \pi r$) divided by the time taken (orbital period $

Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22

Exercises · 7.4

4.22 \times 10^{8}\text{ m}

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Concept understanding — Kepler's Third Law

Kepler's Third Law: The Harmony of the Planets

Imagine you're watching two planets orbiting the Sun. One is close in — Mercury, zipping around in just 88 days. Another is far out — Saturn, taking nearly 30 years to complete one lap. You'd expect the farther planet to take longer, but here's the surprising part: the relationship isn't just "farther = slower." It's much more precise, and it reveals a deep truth about gravity itself.

The Intuition

Think of a planet as a runner on a circular track. The farther out the track, the longer the lap — that's obvious. But Kepler noticed something subtler: if you double the distance from the Sun, the orbital period doesn't just double. It increases by a factor of about $2.8$ (which is $\sqrt{8}$ ). Triple the distance, and the period grows by about $5.2$ (which is $\sqrt{27}$ ).

There's a pattern here. The period seems to grow as the 3/2 power of the distance. Why? Because gravity weakens with distance, so a farther planet feels a weaker pull and moves more slowly — not just because the track is longer, but because it's moving slower along that track.

The Precise Statement

T^2 \propto a^3

For planets orbiting the Sun, if you measure $T$ in Earth years and $a$ in astronomical units (AU, where 1 AU = Earth's average distance from the Sun), the constant of proportionality is exactly 1:

T^2 = a^3

So for Earth: $T = 1$ year, $a = 1$ AU, and $1^2 = 1^3$ — it checks out.

For Mars: $a \approx 1.52$ AU, so $T^2 = (1.52)^3 \approx 3.51$ , giving $T \approx 1.87$ years. That's about 687 days — exactly right.

Note

This law applies to any body orbiting a much more massive central body: moons around planets, satellites around Earth, binary stars around each other. The constant of proportionality changes depending on the mass of the central body.

Why It Works (The Physics)

Newton later showed that Kepler's Third Law is a direct consequence of his law of gravitation. For a circular orbit (a good approximation for most planets), the centripetal force needed to keep the planet in orbit is provided by gravity:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Here $M$ is the Sun's mass, $m$ the planet's mass, $r$ the orbital radius, and $v$ the orbital speed. The speed is related to the period by $v = 2\pi r / T$ . Substituting and simplifying:

\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}

Rearranging:

T^2 = \frac{4\pi^2}{GM} r^3

The quantity $4\pi^2/(GM)$ is a constant for all planets orbiting the Sun. So $T^2 \propto r^3$ — exactly Kepler's law.

Important

The constant $4\pi^2/(GM)$ depends only on the mass of the central body. This means: if you know the period and distance of any moon or planet, you can calculate the mass of the body it orbits. This is how astronomers "weigh" stars, black holes, and galaxies.

A Common Mistake …

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By applying Kepler's Third Law, which relates the orbital period and radius of a satellite to the mass of its central body, we calculate Jupiter's mass using Io's orbital data. Comparing this to the known mass of the Sun reveals that Jupiter's mass is approximately one-thousandth that of the Sun.

To determine the mass of Jupiter and compare it to the Sun's mass, we use Kepler's Third Law of planetary motion. This law, derived from Newton's Law of Universal Gravitation and the concept of centripetal force, provides a direct relationship between the orbital period ( $T$ ) and orbital radius ( $r$ ) of a satellite and the mass ( $M$ ) of the central body it orbits.

The fundamental idea is that the gravitational force exerted by Jupiter on Io provides the necessary centripetal force to keep Io in its orbit. By equating these two forces, we can solve for the mass of Jupiter.

M

›Proof

Derivation of Kepler's Third Law

Consider a satellite of mass $m$ orbiting a central body of mass $M$ in a circular path of radius $r$ .

Gravitational Force: The force of gravity attracting the satellite to the central body is given by Newton's Law of Universal Gravitation:

$F_g = \frac{G M m}{r^2}$
Centripetal Force: For the satellite to maintain a circular orbit, it must experience a centripetal force directed towards the center of the orbit. If the satellite's orbital speed is $v$ , this force is:

$F_c = \frac{m v^2}{r}$
Equating Forces: In a stable orbit, the gravitational force provides the centripetal force:

$\frac{G M m}{r^2} = \frac{m v^2}{r}$
Relating Speed to Period: For a circular orbit, the speed $v$ is the distance traveled (circumference $2 \pi r$ ) divided by the time taken (orbital period $T$ ):

$v = \frac{2 \pi r}{T}$
Substitution and Simplification: Substitute the expression for $v$ into the equated forces equation:

$\frac{G M m}{r^2} = \frac{m}{r} \left(\frac{2 \pi r}{T}\right)^2$

$\frac{G M}{r^2} = \frac{4 \pi^2 r^2}{r T^2}$

$\frac{G M}{r^2} = \frac{4 \pi^2 r}{T^2}$

Rearranging to solve for $M$ :

$M = \frac{4 \pi^2 r^3}{G T^2}$

This is the form of Kepler's Third Law we will use.

Let's proceed with the calculation:

Identify Given Values and Constants:
- Orbital period of Io, $T = 1.769 \text{ days}$
- Orbital radius of Io, $r = 4.22 \times 10^8 \text{ m}$
- Universal Gravitational Constant, $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$
- Mass of the Sun, $M_S = 1.989 \times 10^{30} \text{ kg}$ (This will be used for comparison later).
Convert Units to SI:

The orbital period is given in days, but for consistency with SI units in the gravitational constant, we must convert it to seconds.

$1 \text{ day} = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ seconds}$ .

Therefore,

$T = 1.769 \text{ days} \times 86400 \text{ s/day}$

$T = 152949.6 \text{ s}$
Calculate the Mass of Jupiter ( $M_J$ ):

Now, substitute the values into the formula for $M$ :

M_J = \frac{4 \pi^2 r^3}{G T^2}

M_J = \frac{4 \pi^2 (4.22 \times 10^8 \text{ m})^3}{(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) (152949.6 \text{ s})^2}

Let's calculate the numerator and denominator separately for clarity. …

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Q.Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22×108 m4.22 \times 10^{8}\text{ m}4.22×108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.

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Q.Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 10^{8}\text{ m}$ . Show that the mass of Jupiter is about one-thousandth that of the sun.

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