373
13-7 SATE LLITES: OR B ITS AN D E N E RGY
KEY IDEA
On the launchpad, the ball is not in orbit and thus Eq. 13-40
does not apply. Instead, we must find E 0 ϭ K 0 ϩ U 0 , where
K 0 is the ball’s kinetic energy and U 0 is the gravitational potential energy of the ball – Earth system.
Calculations: To find U 0 , we use Eq. 13-21 to write
ϭ Ϫ4.51 ϫ 10
8
J ϭ Ϫ451 MJ.
The kinetic energy K 0 of the ball is due to the ball’s motion
with Earth’s rotation. You can show that K 0 is less than 1 MJ,
which is negligible relative to U 0 . Thus, the mechanical energy of the ball on the launchpad is
E 0 ϭ K 0 ϩ U 0 Ϸ 0 Ϫ 451 MJ ϭ Ϫ451 MJ.
(Answer)
The increase in the mechanical energy of the ball from
launchpad to orbit is
⌬E ϭ E Ϫ E 0 ϭ (Ϫ214 MJ) Ϫ (Ϫ451 MJ)
ϭ 237 MJ.
(Answer)
This is worth a few dollars at your utility company.
Obviously the high cost of placing objects into orbit is not
due to their required mechanical energy.
ϭ Ϫ
(6.67 ϫ 10
Ϫ11
Nиm
2
/kg
2
)(5.98 ϫ 10
24
kg)(7.20 kg)
6.37 ϫ 10
6
m
U 0 ϭ Ϫ
GMm
R
Sample Problem 13.05 Mechanical energy of orbiting bowling ball
A playful astronaut releases a bowling ball, of mass m
7.20 kg, into circular orbit about Earth at an altitude h of
350 km.
(a) What is the mechanical energy E of the ball in its orbit?
KEY IDEA
We can get E from the orbital energy, given by Eq. 13-40
(E ϭ ϪGMm/2r), if we first find the orbital radius r. (It is
not simply the given altitude.)
Calculations: The orbital radius must be
r ϭ R ϩ h ϭ 6370 km ϩ 350 km ϭ 6.72 ϫ 10
6
m,
in which R is the radius of Earth. Then, from Eq. 13-40 with
Earth mass M ϭ 5.98 ϫ 10
24
kg, the mechanical energy is
ϭ Ϫ2.14 ϫ 10
8
J ϭ Ϫ214 MJ.
(Answer)
(b) What is the mechanical energy E 0 of the ball on the
launchpad at the Kennedy Space Center (before launch)?
From there to the orbit, what is the change ⌬E in the ball’s
mechanical energy?
ϭ Ϫ
(6.67 ϫ 10
Ϫ11
Nиm
2
/kg
2
)(5.98 ϫ 10
24
kg)(7.20 kg)
(2)(6.72 ϫ 10
6
m)
E ϭ Ϫ
GMm
2r
ϭ
ence of the initial circular orbit to the initial period of the orbit. Thus, just after the thruster is fired, the kinetic energy is
ϭ 1.0338 ϫ 10
11
J.
ϭ
1
2 (4.50 ϫ 10
3
kg)(0.96)
2
2p (8.00 ϫ 10
6
m)
7.119 ϫ 10
3
s
2
K ϭ
1
2 mv
2 ϭ
1
2 m(0.96v 0 )
2 ϭ
1
2 m(0.96)
2
2pr
T 0
2
Sample Problem 13.06 Transforming a circular orbit into an elliptical orbit
A spaceship of mass m 4.50 10
3
kg is in a circular Earth
orbit of radius r ϭ 8.00 ϫ 10
6
m and period T 0 ϭ 118.6 min ϭ
7.119 ϫ 10
3
s when a thruster is fired in the forward direction
to decrease the speed to 96.0% of the original speed. What is
the period T of the resulting elliptical orbit (Fig. 13-17)?
KEY IDEAS
(1) The orbit of an elliptical orbit is related to the semimajor axis a by Kepler’s third law, written as Eq. 13-34 ( ϭ
4p
2
r
3
/GM) but with a replacing r. (2) The semimajor axis a
is related to the total mechanical energy E of the ship by
Eq. 13-42 (E ϭ ϪGMm/2a), in which Earth’s mass is M ϭ
5.98 ϫ 10
24
kg. (3) The potential energy of the ship at radius
r from Earth’s center is given by Eq. 13-21 (U ϭ ϪGMm/r).
Calculations: Looking over the Key Ideas, we see that we
need to calculate the total energy E to find the semimajor
axis a, so that we can then determine the period of the elliptical orbit. Let’s start with the kinetic energy, calculating it just
after the thruster is fired. The speed v just then is 96% of the
initial speed v 0 , which was equal to the ratio of the circumferT
2
ϫ
ϭ
Figure 13-17 At point P a
thruster is fired, changing a
ship’s orbit from circular to
elliptical.
r
M
P
13-7 SATE LLITES: OR B ITS AN D E N E RGY
KEY IDEA
On the launchpad, the ball is not in orbit and thus Eq. 13-40
does not apply. Instead, we must find E 0 ϭ K 0 ϩ U 0 , where
K 0 is the ball’s kinetic energy and U 0 is the gravitational potential energy of the ball – Earth system.
Calculations: To find U 0 , we use Eq. 13-21 to write
ϭ Ϫ4.51 ϫ 10
8
J ϭ Ϫ451 MJ.
The kinetic energy K 0 of the ball is due to the ball’s motion
with Earth’s rotation. You can show that K 0 is less than 1 MJ,
which is negligible relative to U 0 . Thus, the mechanical energy of the ball on the launchpad is
E 0 ϭ K 0 ϩ U 0 Ϸ 0 Ϫ 451 MJ ϭ Ϫ451 MJ.
(Answer)
The increase in the mechanical energy of the ball from
launchpad to orbit is
⌬E ϭ E Ϫ E 0 ϭ (Ϫ214 MJ) Ϫ (Ϫ451 MJ)
ϭ 237 MJ.
(Answer)
This is worth a few dollars at your utility company.
Obviously the high cost of placing objects into orbit is not
due to their required mechanical energy.
ϭ Ϫ
(6.67 ϫ 10
Ϫ11
Nиm
2
/kg
2
)(5.98 ϫ 10
24
kg)(7.20 kg)
6.37 ϫ 10
6
m
U 0 ϭ Ϫ
GMm
R
Sample Problem 13.05 Mechanical energy of orbiting bowling ball
A playful astronaut releases a bowling ball, of mass m
7.20 kg, into circular orbit about Earth at an altitude h of
350 km.
(a) What is the mechanical energy E of the ball in its orbit?
KEY IDEA
We can get E from the orbital energy, given by Eq. 13-40
(E ϭ ϪGMm/2r), if we first find the orbital radius r. (It is
not simply the given altitude.)
Calculations: The orbital radius must be
r ϭ R ϩ h ϭ 6370 km ϩ 350 km ϭ 6.72 ϫ 10
6
m,
in which R is the radius of Earth. Then, from Eq. 13-40 with
Earth mass M ϭ 5.98 ϫ 10
24
kg, the mechanical energy is
ϭ Ϫ2.14 ϫ 10
8
J ϭ Ϫ214 MJ.
(Answer)
(b) What is the mechanical energy E 0 of the ball on the
launchpad at the Kennedy Space Center (before launch)?
From there to the orbit, what is the change ⌬E in the ball’s
mechanical energy?
ϭ Ϫ
(6.67 ϫ 10
Ϫ11
Nиm
2
/kg
2
)(5.98 ϫ 10
24
kg)(7.20 kg)
(2)(6.72 ϫ 10
6
m)
E ϭ Ϫ
GMm
2r
ϭ
ence of the initial circular orbit to the initial period of the orbit. Thus, just after the thruster is fired, the kinetic energy is
ϭ 1.0338 ϫ 10
11
J.
ϭ
1
2 (4.50 ϫ 10
3
kg)(0.96)
2
2p (8.00 ϫ 10
6
m)
7.119 ϫ 10
3
s
2
K ϭ
1
2 mv
2 ϭ
1
2 m(0.96v 0 )
2 ϭ
1
2 m(0.96)
2
2pr
T 0
2
Sample Problem 13.06 Transforming a circular orbit into an elliptical orbit
A spaceship of mass m 4.50 10
3
kg is in a circular Earth
orbit of radius r ϭ 8.00 ϫ 10
6
m and period T 0 ϭ 118.6 min ϭ
7.119 ϫ 10
3
s when a thruster is fired in the forward direction
to decrease the speed to 96.0% of the original speed. What is
the period T of the resulting elliptical orbit (Fig. 13-17)?
KEY IDEAS
(1) The orbit of an elliptical orbit is related to the semimajor axis a by Kepler’s third law, written as Eq. 13-34 ( ϭ
4p
2
r
3
/GM) but with a replacing r. (2) The semimajor axis a
is related to the total mechanical energy E of the ship by
Eq. 13-42 (E ϭ ϪGMm/2a), in which Earth’s mass is M ϭ
5.98 ϫ 10
24
kg. (3) The potential energy of the ship at radius
r from Earth’s center is given by Eq. 13-21 (U ϭ ϪGMm/r).
Calculations: Looking over the Key Ideas, we see that we
need to calculate the total energy E to find the semimajor
axis a, so that we can then determine the period of the elliptical orbit. Let’s start with the kinetic energy, calculating it just
after the thruster is fired. The speed v just then is 96% of the
initial speed v 0 , which was equal to the ratio of the circumferT
2
ϫ
ϭ
Figure 13-17 At point P a
thruster is fired, changing a
ship’s orbit from circular to
elliptical.
r
M
P