Mr. Kepple
Name: _______________________
Gravitation Example Problems
Date: ___________ Period: _____
WHERE IS MARS?
Mars’ period (it’s “year”) was first noted by Kepler to be about 687 days (Earth-days). Determine the
mean distance of Mars from the Sun using the Earth as a reference.
Start with Kepler’s 3rd Law, which says
the ratio of 𝑅 to 𝑇 2 is constant for
every planet orbiting the same body.
𝑅
𝑇2
𝐺𝑀
𝜋2
From this we can relate the orbital
radius of Mars 𝑅𝑚 to that of Earth 𝑅𝑒 .
𝑅𝑚
𝑇𝑚2
𝑅𝑒
𝑇𝑒2
Now we simply solve for 𝑅𝑚 and
rearrange so it’s easier to plug the
numbers into a calculator.
const
𝑅𝑒
𝑅𝑚
1/
𝑅𝑚
𝑇𝑚2 𝑅𝑒
𝑇𝑒2
2/
𝑅𝑚
𝑇𝑚
𝑇𝑒
𝑅𝑒
𝑅𝑚
6
65 25
2/
52 𝑅𝑒
The distance of Mars from the
Sun is about 1.52 times
greater than the distance of
Earth from the Sun.
2007 MECH. 2.
In March 1999 the Mars Global Surveyor (MGS) entered its final orbit about Mars, sending data back to
Earth. Assume a circular orbit with a period of 118 minutes
s and orbital speed of
m/s. The mass of the MGS is 930 kg and the radius of Mars is
m.
(a) Calculate the radius of the GS orbit.
Since the orbit is circular the radius of
orbit can be found simply by relating
the orbital speed with the period.
𝑣
2𝜋𝑅
𝑇
Solve for 𝑅 and plugin.
𝑅
𝑣𝑇
2𝜋
𝑅
2π
𝑅
2
m
𝑅≈
m
(b) Calculate the mass of Mars.
Since the orbit is circular
gravity acts centripetally.
Equate the gravitational and
centripetal force.
Solve for 𝑀 and plugin.
AP Physics C – 2013
𝐹𝑔
𝐺𝑀𝑚
𝑅2
𝐹𝑐
𝑚𝑣 2
𝑅
2
𝑀
𝑣 𝑅
𝐺
Gravitation – Page 4
2
𝑀
𝑀
66
−11
6 6 99
2
𝑅 ≈ 66
2
m
m
3/29/14
(c) Calculate the total mechanical energy of the MGS in this orbit.
Total mechanical energy is
just kinetic plug potential.
Substitute formulas for
kinetic and gravitational
potential energy.
𝐸
𝐾+𝑈
𝐸
𝐺𝑀𝑚
𝑚𝑣 2 −
2
𝑅
Substitute for 𝑀 using the
answer from part (b).
𝐸
Now simplify so it’s easier
to plug into the calculator.
𝐸
Oh yeah, total mechanical
energy is just –𝐾!
𝐸
2
2
𝑚𝑣 2 −
𝐸
−
𝐸
2
9
2
−5
9
5
J
2
𝐺𝑚 𝑣 𝑅
𝑅
𝐺
9
𝐸 ≈ −5
J
Don’t forget the negative sign!
𝑚𝑣 2 − 𝑚𝑣 2
− 𝑚𝑣 2
2
(d) If the MGS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new
orbital period of the MGS be greater than or less than the given period?
___ Greater than
___ Less Than
X
Justify your answer.
According to Kepler’s 3rd law, the cube of the radius of orbit is directly proportional to
the square of the period of the orbit. As a result if the orbital radius decreases so too
must the orbital period.
Alternative explanation: The equation from part (b) shows that decreasing the radius
of the orbit leads to an increase in orbital speed. Furthermore, the equation from
part (a) shows that an increase in orbital speed leads to a decrease in orbital period.
(e) In fact, the orbit the MGS entered was slightly elliptical with its closest approach to Mars at
m above the surface and its furthest distance at
6
m above the surface. If the
speed of the GS at closest approach is
m/s, calculate the speed at the furthest point of the
orbit.
Elliptical orbit: use angular
momentum conservation.
𝐿0
The angular momentum of
a point particle is 𝑚𝑣𝑟.
𝑚𝑣0 𝑟0
Since we are given the
distance of the satellite
above the surface we must
add the radius of Mars to
each measurement.
𝑣
𝑣
𝐿
𝑣
+
+
6
𝑣0
𝑚𝑣𝑟
𝑣
𝑟0
𝑣
𝑟 0
𝑅𝑐 + 𝑅𝑚
𝑅𝑓 + 𝑅𝑚
9
𝑣
𝑣≈
𝑣0
2
m/s
m/s