Question

The greater the speed of a satellite, the shorter its orbital period. T/F

Answer 1

TRUE BECAUSE

EXPLAINATION

Kepler's 2^nd Law: A line connecting the Sun and planet sweeps out equal areas in equal times. That is, the orbital speed of any one planet varies inversely with its distance from the Sun (actually, orbital speed varies inversely with the square-root of the distance, but you needn't worry about that detail).

This one you can understand conceptually, once you understand Newton's Laws of Motion and Law of Gravity. If the planet changes its distance from the Sun as it orbits, then the force of gravity between them must change. If the force that the Sun exerts on the planet increases (as the planet moves closer), then the acceleration of the planet must increase, resulting in a higher orbital speed, and vice versa.

See? You don't need a single equation! Let's review where this comes from, referring back to a couple of equations only as reminders of the relationships between physical concepts:

For any force on any mass: F = m x a; a is the acceleration of mass m (Newton's 2^nd Law of Motion).

The force of gravity between 2 bodies and Newton's 2^nd Law of Motion:
F_grav = M₁ (G M₂ / d²) = M₁ a₁ = M₂ (G M₁ / d²) = M₂ a₂

Since the distance between the Sun's center and the planet's center (d) changes, the gravitational force between them does too. Comparing the expressions for gravitational force with the general force law (Newton's 2^nd Law of Motion), one can see that acceleration of M₁(let's say the Sun) is G M₂ / d², and that of M₂(the planet) is G M₁ / d². That is, the acceleration of a planet in its orbit around the Sun depends upon the mass of the Sun and the inverse square of the planet's distance from the Sun. As the planet moves further away in its orbit around the Sun, the gravitational force exerted by the Sun on the planet decreases. If the force exerted on the planet decreases, the planet's acceleration, proportional to M_sun/d², must also decrease, resulting in a lower orbital speed. We won't worry about how orbital acceleration is converted into orbital speed in this class. Just think of accelerating a bowling ball down the bowling alley - a ball undergoing a greater acceleration from rest (in your hands) attains a higher speed moving down the lane. That's all you need to understand, and you already knew that.

The two very important conservation laws in nature: that of angular momentum (r x m x v = constant) and energy (here: kinetic + gravitational potential = constant) also explain Kepler's 2^nd Law. In the first case, as the planet's distance from the Sun (here called 'r') decreases, it's orbital speed must increase in order that its orbital angular momentum (r x m x v) remains constant. Just the opposite occurs as the planet's distance increases from the Sun. In the second case, as the planet's distance from the Sun decreases, so does its gravitational potential energy. Thus the planet's kinetic energy (proportional to the square of its orbital speed) must increase to compensate to keep the total energy = sum of kinetic + potential energies constant (conservation of energy). And of course, just the opposite applies as the planet's distance increases from the Sun - its potential energy increases, thus its kinetic energy (and so orbital speed) must decrease.

Kepler's 3^rd Law: p²(yrs) = a³(AU), meaning larger orbits (a) have longer orbital periods (p), and the average orbital speeds are slower for planets with larger orbits.

After deriving gravitational orbital motion from the laws of motion, law of gravity and quite a bit of calculus, Newton found that Kepler's 3rd Law should actually be this: p²(yrs) = a³(AU) / (M₁ + M₂); where the masses are measured in units of our Sun's mass. Newton also showed that this equation may also be expressed in physical units as
p² = 4pi²a³/ G(M₁ + M₂)
for p measured in seconds, a in meters, and M in kg (and pi = 3.141592653...).

How can we understand this relation between the period of the orbit, mean distance, and sum of masses? Just as we did above. Recalling the underlined statement above, the acceleration of a more distant planet due to the force of gravity between it and the Sun must be smaller than the acceleration of a planet near to the Sun. Result: a more distant planet must orbit the Sun at a lower average orbital speed. Slower orbital speed and a larger orbit mean that its orbital period must be longer! No equations!

Important: The force of gravity between the Sun and a planet can be larger than that between the Sun and Earth, yet that planet's acceleration can be either larger (if closer to the Sun) or smaller (if further from the Sun) than Earth's accleration in its orbit around the Sun. This is because the force depends upon the product of both masses and the square of the distance between them, whereas the planet's acceleration depends on just the mass of the Sun and the square of the distance between the Sun and planet. For example, the force between Jupiter and the Sun is 11.7x that between Earth and the Sun, and yet Jupiter's orbital acceleration is a factor of 27 smaller than Earth's (because Jupiter lies 5.2x further away from the Sun than Earth and 5.2 x 5.2 = 27). Review the discussion of Kepler's 2^nd Law, above.

Back to the 3^rd law. Note that the relationship between the orbital period and the average separation between the masses actually depends upon the sum of the two masses orbiting the common center of mass. That this relationship should involve mass should make some sense given what you know about the relationship between force, mass, and acceleration, and that the force of gravity itself depends upon masses of the bodies involved. Surely, you can imagine that if the star that Earth orbited were more massive than our Sun, Earth's acceleration would be greater, it's orbital speed faster, and so its orbital period shorter.

Note that for the case of our Solar System's planets orbiting the Sun, the sum of the masses correction to Kepler's 3^rd law (M₁ + M₂, measured in units of our Sun's mass) amounts to a factor of at most 1.000955 (for the Sun + Jupiter = 1 + 1/1047). Thus Kepler had it nearly right in his description of the orbits of the objects around our Sun. On the other hand, this mass correction term is important, i.e., significantly different from a factor of 1, for all other cases involving the orbits of two masses when one of them is not our Sun (e.g., Earth-Moon, star-star, or even planet-satellite).

So does the mass of the planet have a significant impact upon its orbital period (or orbital speed) about some star? Given that planets are by definition almost always much less massive than the stars they orbit, the practical answer is "NO." Is there anyeffect? Well, yes of course there is, since the force of gravity does consider the masses of pairs of objects. Look at Newton's corrected version of Kepler's 3^rd law again. It does indeed include the masses of both objects. What if Jupiter with its mass of 318 Earths orbited the Sun at Earth's location rather than Earth (ignoring the presence of nearby Mars and Venus)? Well, then the mass correction factor would be 1.000955 instead of that of 1.000003 for Earth. In that case the orbital period of Jupiter in Earth's orbit would be a tiny, tiny amount shorter (1.000476 times shorter). How the sum of the two masses comes into Kepler's 3^rdlaw delves into physical concepts that are beyond this class.