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Physics Unit 6 Lecture Notes
According to Newton’s first law of motion, an object with no unbalanced forces acting on it will
continue moving in a straight line at a constant speed. Would a planet with no forces acting on it
obey Kepler’s second law of planetary motion (the law of equal areas swept out in equal times)?
In the diagram below, the planet is traveling at a constant velocity so that it travels the same
distance, ∆
∆ , for several successive time intervals. The letter is at the position of the
Sun.
𝑣∆𝑡
𝑣∆𝑡
𝑣∆𝑡
𝑣∆𝑡
𝐴
𝐵
𝐶
𝐷
𝐸
In other words, ̅̅̅̅ ̅̅̅̅ ̅̅̅̅ ̅̅̅̅
∆ .
Since the height is the same for all of the triangles,
(∆
)
(∆
)
(∆
)
(∆
)
.
So the answer is yes, a planet with no unbalanced forces acting on it obeys Kepler’s second law.
𝑆
Now suppose that a Sun-directed force, , acts very briefly on the planet as it is passing point .
(By “Sun-directed,” it is meant that the force vector is parallel with the line segment ̅̅̅̅.) The
force will cause an acceleration and, therefore, a change in the velocity of the planet. Because
, and ∆
∆ ,∆
. Now, instead of traveling the path ̅̅̅̅
∆ during the
′
′
next time interval, the planet will travel ̅̅̅̅̅ ⃑⃑⃑ ∆
∆
∆ ∆ . (See diagram.)
Since we have already established that
(∆
)
(∆
) , all that is needed
to show that Kepler’s second law is still obeyed is
′ ).
(∆
)
(∆
to show that
But the two triangles share the same base, ̅̅̅̅ , and
′
̅̅̅̅ (because ∆
the fact that ̅̅̅̅̅
) means
that the two triangles have the same height.
Therefore, we can conclude that the two triangles
have the same area, and Kepler’s second law
holds.
𝐹
𝐴
𝑣∆𝑡
𝐵
𝑣∆𝑡
𝐶
∆𝑣∆𝑡
𝑣 ′ ∆𝑡
𝐶′
𝑆
Here, though is the key point to consider: if the
′ would not be parallel to ̅̅̅̅ , the
force had been in any direction other than towards the Sun, then ̅̅̅̅̅
′
triangles ∆
and ∆
would not have equal areas, and Kepler’s second law would not hold.
From this, we (and Newton) can conclude that Kepler’s second law can be true if, and only if,
the force that keeps the planets in orbit is always in the direction of the Sun.
Physics Unit 6 Lecture Notes
We will now look for a rule that governs the size of the force on the planet. In order to
accomplish this without inventing the calculus, we will have to work with circular orbits. While
it is true that Kepler’s first law is that planets orbit on elliptical paths with the Sun at one focus, it
is also true that a circle is an ellipse, and that most planets’ orbits are nearly circular.
According to Newton’s second law, the force on the planet is equal to the mass of the planet
multiplied by its acceleration (
). Since the acceleration is
(
centripetal, we can rewrite this as
).
The problem with this formula is that it says that the force that the Sun exerts on the planet at this
moment depends on how long it will take for the planet to complete its orbit. This may be a valid
empirical relationship, but it cannot give any insight into cause and effect. We need a formula for
the force that depends only on the physical situation of the moment. This is where Kepler’s third
law (
) comes in. We rearrange it and replace
the force equation with this new expression:
(
with :
)
(
. Then replace the
in
)
This shows that the force exerted by the Sun on a planet is inversely proportional to the square of
the distance between the Sun and the planet. This turns out to apply equally well to a planet on
an elliptical orbit. In that case, though, the force is continually changing as the distance between
the Sun and the planet changes.
The next step requires a philosophical or metaphysical step. Newton took the position that the
laws of physics must apply equally to all objects. Therefore, the force between two objects
cannot be proportional to only one of the objects. If it is proportional to the mass of one of the
objects, it must be proportional to both masses. Ignoring the constants in our previous equation
for a moment, we can express this relationship as follows:
( )
To express this as an equation, we need only to add in a constant of proportionality, which we
will symbolize with a capital G:
( )
We now arrive at the last step. The Sun exerts a force on the planets; planets exert forces on their
moons; the Earth exerts forces on every massive object on its surface. Newton makes the
assertion finally that every object in the universe that has mass exerts a force on every other
massive object. This is Newton’s Law of Universal Gravitation:
where
is the distance between the centers of object A and object B.
Physics Unit 6 Lecture Notes
At this point, we should go back and look to see what happened to Kepler’s constant, k. Looking
at what disappeared from one equation and what appeared in the next, we can conclude that,
or
This explains why the value for k depends on the object being orbited. For the moons of Jupiter,
for example,
.
Physics Unit 6 Lecture Notes
Problem-solving:
1. If the problem gives you the radius and period of orbit for one moon of a planet, and the radius
of orbit for another moon, and asks you for the period of orbit for the second moon, use the
equation for Kepler’s third law, keeping the units given in the problem:
2. If the problem gives you the radius and period of orbit for one moon of a planet, and the
period of orbit for another moon, and asks you for the radius of orbit for the second moon, use
the equation for Kepler’s third law, flipping the equation over and keeping the units given in
the problem:
3. If the problem describes an object orbiting the Sun and gives the orbital period in years, and
asks for the orbital radius in A.U., or gives the orbital radius in A.U. and asks for the orbital
period in years, use the equation for Kepler’s third law along with the values for Earth’s orbit
in A.U. and years:
3. If the problem gives you the masses of two objects and the distance between them and asks for
the gravitational force between them, use Newton’s law of universal gravitation.
4. If the mass and radius of a planet in proportion Earth values, and asks for the weight or
acceleration due to gravity on the other planet, write out the appropriate equation and remove
the factors to the front. Then multiply the factors by the Earth value.
(
(
)
)
(
(
)
)
or
(
(
)
)
(
)
(
)
5. If the problem gives you orbital information a moon and asks you for the mass of the planet
that the moon orbits, use Newton’s second law of motion, substitute the gravitation formula
for net force, the mass of the moon for the mass, and the centripetal acceleration expression
for acceleration:
(
Solve for the mass of the planet.
)