Gravitational Interaction II
Level : Physics I
Teacher : Kim
Satellites in Circular Orbit
The satellites in orbit around Earth are kept on its circular path by a centripetal force, which is the
gravitational force.
There is only one speed that a satellite can have if the satellite is
to remain in an orbit with a fixed radius. Any other speed will
cause the satellite to escape orbit or crash back toward the
planet
r
v
satellite!
Fg
Since the gravitational force is acting on the satellite of mass m
in the radial direction(=toward the center), it alone provides the
centripetal force.
𝐺𝑚𝑀𝑒
𝑚𝑣 2
𝛴𝐹𝑟 =
=
𝑟2
𝑟
-11
2
2
where G=6.67×10 N·m /kg is the universal gravitational constant, Me is the mass of the Earth,
and r is the distance from the center of the earth to the satellite. Solving for v of the satellite gives
𝐺𝑀𝑒
𝑟
𝑣=√
: orbital speed
If the satellite is to remain in an orbit of radius r, the speed must have precisely this value.
The period T of a satellite is the time required for one orbital revolution. In a uniform circular
motion, the period is related to the speed of the motion by
𝑣=
𝐺𝑀𝑒
,
𝑟
Since orbital speed is 𝑣 = √
2𝜋𝑟
𝑇
: orbital speed
then
𝐺𝑀
2𝜋𝑟
√ 𝑒=
𝑟
𝑇
Solving for T gives
𝑇=
where the standard unit for period is seconds
2𝜋𝑟 3/2
√𝐺𝑀𝑒
𝑣=
2𝜋𝑟
𝑇
𝑣=√
𝐺𝑀
𝑟
𝑇=
2𝜋𝑟 3/2
√𝐺𝑀
Q1) If a satellite in orbit increases its speed, the satellite will
a) explode
b) get bigger c) fall back to surface
d) break orbit
e) turn to stone
Q2) If a satellite in orbit wants to maintain orbit farther away from the surface, the satellite must
a) increase its speed
b) increase its mass
c) decrease its speed
d) change color
Q3) Determine the speed of the Hubble space telescope orbiting at a height of
h=589km(1km=1000m) above the earth’s surface? (earth’s mass Me=5.98×1024kg, earth’s radius
Re=6.38×106m)
a) 7.56×103 m/s
b) 4.25×103 m/s
c) 2.18×103 m/s
d) 1.05×103 m/s
h
R
Q4) A satellite needs to maintain orbit 6×105m above the surface of Jupiter. Jupiter has a mass of
1.9×1027kg and a radius of 7.14×107m. What must be speed of the satellite to have a stable orbit
at the desired height?
a) 1.2×104 m/s
b) 2.25×104 m/s
c) 4.2×104 m/s
d) 5.2×104 m/s
Q5) The satellite circles the earth in an orbit whose radius is twice the earth’s radius. The earth’s
mass is 5.98×1024kg, and its radius is 6.38×106m. What is the period of the satellite?
a) 1.44×104 s
b) 2.34×104 s c) 4.06×104 s
d) 5.28×104 s
Q6) The earth travels around the sun once a year at a distance of 1.50×1011m. From these data
determine (i) the mass of the sun and (ii) the orbital speed of the earth
a) 2×1030kg, 3×104m/s
b) 2×1028kg, 3×102m/s
30
4
c) 2×10 kg, 1.2×10 m/s
d) 2×1028kg, 1.2×102m/s
*~1 year=365days×24hours×3600seconds
𝑣=√
𝐺𝑀
𝑟
𝑇=
2𝜋𝑟 3/2
√𝐺𝑀
Escape Speed
If someone holds you from the back, you need to a certain amount of force to break free.
If a rocket wanted to break free from the gravitational pull from
the Earth, then the rocket must thrust a certain amount of force to
escape the Earth
The formula to find the escape speed from any planetary body is
expressed as
2𝐺𝑀
𝑣𝑒 = √
𝑅
where the universal gravitational constant G=6.67×10-11, M is the mass and R is the radius of the
planetary body
*~ Note that more speed is needed to escape from a planetary body compared to just orbiting it.
𝐺𝑀𝑒
𝑟
𝑣=√
2𝐺𝑀
𝑅
< 𝑣𝑒 = √
Example1) Find the escape speed to break free Earth’s gravity and leave Earth. The Earth’s mass
is Me=6×1024kg, radius Re=6.38×106m.
Q7) Some asteroids have such small mass, your running speed is sufficient to run yourself into
orbit. i) If your top speed is 8m/s(=17.9mph), which asteroid is the best candidate to run into orbit
so that you will maintain a stable orbit 100m above the surface?
See table next page for mass and radius
a) Thebe
b) Mathilde
c) Phobus
d) Titan
ii) If you are parked on the asteroid Thebe, what is minimum speed needed to ‘escape’ Thebe?
(To convert m/s into mph, multiply the speed(m/s) by 2.232)
The following list is most of the planetary bodies(except Earth) in our solar system, from the Sun
to planets, dwarf planets and asteroids.
Sun and Planets
Sun
Mercury
M = 1.99×1030kg
R = 6.96×108m
M = 3.3×1023kg
R = 2.44×106m
Venus
Mars
M = 4.87×1024kg
R = 6.05×106m
M = 6.42×1023kg
R = 3.40×106m
Jupiter
Saturn
M = 1.90×1027kg
R = 7.15×107m
M = 5.68×1026kg
R=6.03×107m
Uranus
Neptune
M = 8.68×1025kg
R = 2.56×107m
M = 1.02×1026kg
R = 2.48×107m
Moons, Dwarf Planets and Asteroids
Earth’s moon
Deimos
(Mars’ moon)
M = 7.35×1022kg
R = 1.74×106m
M = 1.48×1015kg
R = 6.2×103m
Phobus
(Mars’ moon)
Titan(Saturn’s largest
moon)
M = 1.07×1016kg
R = 1.11×104m
M = 1.35×1023kg
R = 2.58×106m
Ganymede(Jupiter’s
largest moon)
Thebe(one of Jupiter’s
smaller moon)
M = 1.48×1023kg
R = 2.63×106m
M = 4.3×1017 kg
R = 4.93×104m
Ceres
(Largest asteroid)
Haumea
(dwarf planet)
M = 9.43×1020kg
R = 4.87×105m
21 Lutetia
(asteroid-belt)
M = 4.00×1021kg
R = 7.18×105m
443Eros
(Near-Earth asteroid)
M = 1.7×1018 kg
R = 6.00×104m
Itokawa(Near-Earth
asteroid)
M = 3.58×1010 kg
R = 270m
M = 6.69×1015 kg
R = 1.72×104m
Mathilde
(asteroid-belt)
M = 1.03×1017 kg
R = 2.69×104m