Problem_Set_#12
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School
Bergen Community College *
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Course
101
Subject
Astronomy
Date
Apr 3, 2024
Type
docx
Pages
3
Uploaded by MagistrateFlag4534 on coursehero.com
Anthony Putrino
Problem Set # 12- Problems 7, 20, 30, 36, 47
7.
At new moon, the Earth, Moon, and Sun are in a line. Find the direction and magnitude
of the net
gravitational force exerted on
(a)
the Earth,
(b)
the Moon, and
(c)
the Sun.
Solution: Add all the forces to find direction and magnitude. (a)F
E = G(M
E
M
S
)/r
2
E-S
+G(M
E
M
M
)/r
2
E-M →GM
E
((M
S
/r
2
E-S
)+M
M
/r
2
E-M
) = (6.67*10
-11
N*m/kg
2
)(5.97*10
24
kg)((2*10
30
kg/((1.50*10
11
m)
2
)+(7.35*10
22
kg)
((3.84*10
8
m)
2
) = 3.56*10
22 N, Direction towards the Sun (b) F
M = G(M
S
M
M
)/r
2
S-M
-G(M
E
M
M
)/r
2
E-M →GM
M
((M
S
/r
2
S-M
)-M
E
/r
2
E-M
) =(6.67*10
-11
N*m/kg
2
)(7.35*10
22
kg)((2*10
30
kg/((1.50*10
11
m)
2
)-3.84*10
8
kg)-
(5.97*10
24
kg)/ (3.84*10
8
m)
2
) =2.40*10
20 N, direction towards the Sun
(c) F
S = -G(M
S
M
M
)/r
2
S-M
-G(M
S
M
E
)/r
2
S-E →-GM
S
((M
M
/r
2
S-M
)+M
M
/r
2
S-E
) =(6.67*10
-11
N*m/kg
2
)(2*10
30
kg)((7.35*10
22
kg/((1.50*10
11
m)
2
-
3.84*10
8
m)
2
+(5.97*10
24
kg) ((1.50*10
11
m)
2
) = 3.58*10
22 N, direction towards the Earth/Moon 20.
At some point along the direct path from the center of the Earth to the center of the
Moon, the gravitational force of attraction on a spacecraft from the Moon becomes
greater than the force from the Earth.
(a)
How far from the center of the Earth does this
occur?
(b)
At this location, how far is the spacecraft from the surface of the Moon? How
far is it from the surface of the Earth?
Solution: (a) Solve by setting the Force of the Earth to the Force of the Moon and solve for r. G((m
s
m
E
)/(r
2
))=G((m
s
m
m
)/(R-r
2
)) →m
E
(R-r)
2 = m
M
r
2 →R-r = sqrt(m
M
/=m
E
r) →r = R/(1+
(sqrt(m
M
/m
E
) = (3.84*10
8
m)/1+sqrt(7.35*10
22
kg)/(5.97*10
24
kg) = 3.46*10
8
m
(b) Solve by subtracting the distance of the spacecraft with the radius of the Moon and then the Earth. Distance of spacecraft from the center of the Moon: r = R-r
= 3.84*10
8
m - 3.46*10
8
m = 3.8*10
7
m
Subtract from the radius of the Moon: = 3.8*10
7
m - 1.74*10
6
m = 3.6*10
7
m Subtract from the radius of the Earth: = 3.8*10
7
m - 6.37*10
6
m = 3.40*10
8
m
30.
GPS (Global Positioning System) satellites orbit at an altitude of
2.0×10^7
m
.
Find
(a)
the orbital period, and
(b)
the orbital speed of such a satellite.
Solution:
(a) Solve for T.
T = (2pi/sqrt(GM
Earth
))r
(3/2)
= (2pi/sqrt(6.67*10
-11
N*m/kg
2
)(5.97*10
24
kg))*(2.0*10
7
+6.37*10
6
m)
3/2 = 43000 seconds (b) Solve by dividing circumference by period. v = 2pir/T = 2pi(2.637*10
7
m)/4.3*10
4
sec = 3900m/s 36.
The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957.
The mass of Sputnik I was 83.5 kg, and its distances from the center of the Earth at
apogee and perigee were 7330 km and 6610 km, respectively. Find the difference in
gravitational potential energy for Sputnik I as it moved from apogee to perigee.
Solution:
Solve for ΔU
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