Problem Set 12
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SUNY Empire State College *
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Course
1200
Subject
Astronomy
Date
Feb 20, 2024
Type
docx
Pages
4
Uploaded by DeaconMosquitoMaster1223
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.
To find the direction and magnitude of the net gravitational force of sun, moon, and earth.
It would follow as:
(6.67x 10^-11N*m/kg2) (5.97x10^24kg) ((2x10^30kg/((1.50x10^11m)2) +
(7.35x 10^22kg) ((3.84x 10^8m) ^2)
________
= 3.56 x 10^22 N
(direction towards sun)
________
(6.67 x 10^-11N*m/kg2) (7.35 x 10^22kg) ((2 x 10^30kg/ ((1.50 x 10^11m)2)-3.84 x 10^8kg)
-(5.97 x 10^24kg)/ (3.84 x 10^8m) ^2) ________
=2.40 x 10^20 N
(direction towards sun)
________
(6.67x10^-11N*m/kg2)(2x10^30kg)((7.35x10^22kg/((1.50x10^11m)^2-
3.84x10^8m)^2+(5.97 x 10^24kg)((1.50 x 10^11m)^2) ________
= 3.58 x 10^22 N
(direction towards 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?
a) r1 = 9.0139 (D -r1)
r1 (1+ 9.0139) = 9.0139 D
r1 = 9.0139 / 10.0139 D
r1 = 0.9001 3.84 10^8
r1 = 3.46 x 10^8 m
b)
Given:
Radius of moon = 1.74 x 10^6m
Distance of moon = r2 = D-r1
3.84 x 10^8 - 3.46 x 10^8
= 3.8 x 10^7m
Now the distance measured from surface for the spacecraft is:
3.8 x 10^7 – 1.74 x 10^6 = 3.6 x 10^7m
c) now for the distance of earth’s surface:
3.46 x 10^8m – 6.37 x 10^6m
= 3.40 x 10^8m
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.
a)
T = 2pi sqrt (r^3 / GM)
Mass(earth)= 5.97 x 10^24
Gravitational = 6.67 x 10^-11 N*m /kg
T = (2pi sqrt (6.67 x 10^-11 N*m / kg) (5.97 x 10^24kg)) x (2.0 x 10^7 + 6.37 x 10^6m) ^3/2
= 43000 seconds
The orbital speed is 43000 seconds.
b)
Now it will follow as v=2pi x r / T
2pi (2.64 x 10^7m)/ 4.3 x 10^4 seconds = 3900 m/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.
gravitational potential energy (U) = - G x m x M/r
G is the gravitational constant (6.67×10^−11 N*m^2/kg^2
),
m is the mass of the object (83.5
kg),
M is the mass of the Earth (5.972×10^24
kg),
r is the distance between the center of the Earth and the object.
(6.67×10−11
N*m/kg)
x (83.5
kg) x (5.972×10^24
kg)/7,330,000m
+ (6.67×10^−11
N*m/kg)
x (83.5
kg)
x (5.972×10^24
kg)
/6,610,000m = -4.94 x 10^8 Joules
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47.
Several meteorites found in Antarctica are believed to have come from Mars, including
the famous ALH84001
meteorite that was once thought to contain fossils of ancient life
on Mars. Meteorites from Mars are thought to get to Earth by being blasted off the
Martian surface when a large object (such as an asteroid or a comet) crashes into the
planet. What speed must a rock have to escape Mars?
Given:
v is the escape velocity,
G is the gravitational constant (6.67×10^−11 N*m/kg),
M is the mass of Mars (6.4171×10^23 kg)
R is the radius of Mars = 3390km
v= sqrt(2GM/R)
sqrt (2(6.67×10^−11 N*m^2/kg^2) x (6.4171×10^23 kg) / (3390 x 10^23) = 5.03km/s is the
speed the rock must have to escape mars.