Name _________________________
Date: ________
Planetary Properties - Orbits
Purpose: To familiarize students with the various components that determine a planet’s
movement around the Sun.
Goals: Upon successful completion of this lab assignment, students will be able to:
1. Use Newton’s Law of Gravity to calculate the forces between the Sun and the
planets.
2. Understand the physical significance of planetary properties.
3. Understand the dependency between certain planetary properties.
4.
Newton’s Law of Gravity
Isaac Newton (1642 - 1727) first proposed the idea that objects fall to the ground
because of the force of gravity. Gravity, Newton says, is an inherent property of any
massive object. As long as an object has a mass, it will pull on other objects with mass.
Gravity, like all other forces, is a vector quantity in that we need to know the magnitude
(i.e. amount) of the force as well as the force’s direction. Luckily, the direction of the
gravitational force is always the same: toward the center of the objects in question. The
magnitude of the force can be calculated by using the Universal Law of Gravitation:
FG =
Gm1m2
, where
r2
FG  magnitude of the force of gravity (N)
m1, m2  mass of object 1 and object 2 (kg)
 between the two objects (m)
r  separation
G  Universal Gravitational Constant
= 6.67 x 10 -11 (
N  m2
)
kg 2
Example:
What is the magnitude of the force of gravity between the Earth and the Sun?

To calculate the force of gravity between the Sun and Earth we will need to know the
following information, which is available in a textbook or on an appropriate website:
m1  mass of the Sun = 1.99 x 10 30 kg
m2  mass of the Earth = 5.96 x 10 24 kg
r  1.5 x 1011 m

To calculate the magnitude of the force of gravity, simply substitute the values in the
appropriate places in the Universal Law of Gravitation and then follow through with the
calculation.
Gm1m2
FG =
=
r2
6.67 x 10 1.99 x 10 5.96 x 10 
1.5 x 10 
-11
30
11 2
24
=
7.91 x 10 44
2.25 x 10 22
FG = 3.52 x 10 22 N

According to Newton’s 3rd Law of Motion, the force of 3.52 x 1022 N is felt equally
by both the Sun and the Earth, but in opposite directions. The Sun is being pulled toward
 the center of the Earth with 3.52 x 1022 N while the Earth is being pulled toward the
center of the Sun with the same force. How each object reacts to the force is different
and can be understood by looking at the Law of Inertia. Since the Sun is much more
massive than the Earth, the Sun will resist this force more intently than the Earth. As
such, it is adequate to visualize the Earth being held by the force of gravity in a closed
orbit around a nearly motionless Sun.
1. Use the Universal Law of Gravitation to calculate the force of gravity between the
Sun and Jupiter. Show the steps in your calculation; report your answer in
scientific notation.
2. Use a ratio to calculate how many times greater the force of gravity between the
Sun and Jupiter is than the force of gravity between the Sun and Earth. Show the
steps in your calculation; report your answer in scientific notation.
According to Newton’s 2nd Law of Motion (the Law of Inertia), a planet that is
moving at some velocity (v) will do so in a straight path so long as an outside force
doesn’t act on it. However, planets are under the influence of the force of gravity brought
on by the presence of the Sun’s mass. This centralized force causes the planet to be
pulled (i.e. accelerated) toward the Sun’s center (Figure 7.3). This combination of the
planet’s desire to wander a straight path through space with its experiencing centripetal
acceleration, the planet ends up being trapped in a closed orbit around the Sun.
Velocity
Planet
Acceleration
Sun
Resulting Orbital Motion
Figure 1.1: Orbital motion of a planet around the Sun.
Orbital Properties
Planets move around the Sun in closed paths, referred to as orbits. Certain properties
of a planet’s orbit can affect the probability of whether or not life will develop. The
properties of orbital motion have been well understood ever since Johannes Kepler (1571
- 1630) first proposed his three laws of planetary motion nearly four centuries ago. In
this section we will investigate each of Kepler’s laws of planetary motion in order to
better understand how spacecraft are sent to other worlds in our solar system.
Kepler’s 1st Law: Shapes of Orbits
Kepler’s first law identifies the shape of the orbital path of a planet as it travels
around the Sun is that of an ellipse and not a circle. An ellipse differs from a circle in
that it has two focal points (foci) that are both the same distance from the center, whereas
a circle has only one focal point located on the exact center (see Figure 1.2). This means
that for an ellipse, there are axes of many different sizes that can be drawn through the
center. There are two important axes that are important references for an ellipse. The
minor axis is the shortest axis that can be drawn across the ellipse through the center
while the major axis is the longest.
Foci
Focus
Ellipse
Figure 1.2: Comparison of an ellipse to a circle.
Circle
Kepler also noted that the Sun is located at one of the focal points of a planet’s
elliptical orbit. This means that the distance between the Sun and a planet is always
changing as the planet moves around its orbit. As a standard way of comparing orbit
sizes, the average Sun-planet distance is always referenced. Below are several important
properties to know when discussing planetary orbits:
Semi-major axis (a): half the length of the major axis; this distance is equivalent to
the average Sun-planet distance.
Eccentricity (e): defines the “shape” of the orbit by taking the ratio of the Sun
center distance (c) to the semi-major axis (a){e = c/a}. See Figure 7.5.
Periapsis: the point in an orbit that is closest to the body being orbited.
Apoapsis: the point in an orbit that is farthest from the body being orbited.
Periapsis and apoapsis are general terms used when discussing orbits around an
unspecified body. More specific terms are used when the body being orbited is known.
For example, the terms perihelion and aphelion are used to refer to the point in a planet’s
orbit that is closest and furthest from the Sun, respectively.
c
a
Sun
c = 1.6
a = 3.5
e = c/a = 1.6/3.5 = 0.46
Figure 1.3: Determining the eccentricity (e) of a planetary orbit.
Kepler’s 2nd Law: Equal Area in Equal Time
Kepler’s Second Law of Planetary Motion states that an imaginary line connecting
the Sun and planet will “sweep out” equal areas within the orbital plane within equal
units of time. Though this is strictly an observation of planetary behavior, looking at this
discovery in terms of Newton’s laws reveals a deeper significance.
As a planet moves along its elliptical orbit, the distance separating the planet and the
Sun grows and shrinks over time. This change in the separation means that the force of
gravity between the two also changes. The result is that a planet will end up moving
faster in its orbit when it is closer to the Sun and slower when it’s further away.
Kepler’s 3rd Law: P2 = a3
Kepler’s Third Law of Planetary Motion draws a direct correlation between the
orbital period of a planet (measured in years) to its average distance from the Sun
(measured in AU). Though the accuracy of this law is improved upon using Newton’s
laws, Kepler’s third law in its stated form will be useful enough for our investigations.
Example: The orbital period of Venus is measured to be 0.6152 years. What is the
average distance between the Sun and Venus?
Solution:
2
Porb
= a3
a 3 = (0.6152) 2 = (0.3784)
a =
3
(0.3784)
a = 0.7233 AU

Use the orbit of a fictitious planet orbiting a generic star in Diagram 1.1 to answer
Questions 3 – 11. Show the steps in each calculation; report your answer in scientific
notation where appropriate.
Star
Planet
Diagram 1.1: Orbit of a fictitious planet orbiting a star.
3. Draw in the major axis of the planet’s orbit.
4. Mark the center of the major axis with an ‘x’.
5. Use a ruler to determine the length of the semi-major axis and the Sun-center
distance.
Semi-major axis (a) = ________ cm
Sun-center distance (c) = ________ cm
6. Calculate the eccentricity (e) of the orbit.
e = ________
7. Label the point of perihelion and aphelion on the planet’s orbit.
8. If the scale of Diagram 1.1 is, 1 cm = 3 AU, determine the planet’s true average
distance from the star it is orbiting.
True average distance = ________ AU
9. Between which two planets would this planet be located if it were orbiting within
our solar system?
10. Calculate the orbital period for the planet.
Porb = ________ yrs
11. Where will the planet be when it is moving with its greatest velocity?