Question #8
Angular momentum refers to the resistance of a rotating body to change its spin rate or direction of spin. Angular momentum can be determined by the product of the moment of inertia and the angular velocity, or by the cross product of the moment arm and the product of the mass and velocity vector of the body (Sellers 2004, 112-113). For this particular problem, we will use the latter to determine the angular momentum.
H=R×mV
H=0.5 m×0.25 kg2 ms
H=0.25 kg∙m2s
Since angular momentum is conserved, the angular momentum of the system every point in time after the person releases the mass will remain constant. Note that the mass will travel tangentially to the release point, and will continue to travel in a straight line. Therefore, the velocity vector, as well as the moment of arm, will change accordingly. To be specific, as the mass continue to travel, the velocity vector will decrease as the moment of arm increases. In this way, the angular momentum is conserved.
Question 10
According to Newton’s Law of Universal Gravitation, the gravitational attraction between two objects depends on their masses and the distance between their centers of mass (Sellers 2004, 116).
Fg=Gm1m2R2
In this equation, G refers to the universal gravitational constant (6.67×10-11 N∙m2kg2), m1 and m2 refer to the masses of the bodies, and R refers to the distance between their centers of mass. Let us use this equation to solve this problem.
Fg=Gm1m2R2
Fg=6.67×10-11 N∙m2kg21×106 kg8×106 kg100 m2
Fg=5.336×10-2 N
Question #11
At the surface of the Earth, the acceleration due to gravity is around 9.8 m/s2. In general, the acceleration due to gravity applied to an object changes depending on the mass of the other body and the distance of the object to the other body (Sellers 2004, 118).
ag=μEarthR2
In this equation, μEarth refers to the product of the mass of the earth and the universal gravitational constant, and the value is equal to 3.986×1014 m3s2.
ag=μEarthR2
ag=3.986×1014 m3s225000 m+6378137 m2
ag=9.722 ms2
Question #12
Provided that the mechanical energy is conserved, the sum of the potential energy and the kinetic energy of the system must remain constant throughout the time frame (Sellers 2004, 124).
E=PE+KE
In this case, let us consider two stages, the first one being the baseball on the Empire State building, and the second when the baseball hits the ground. According to the conservation of mechanical energy, the sum of the potential energy and the kinetic energy of the system during the first stage must be equal to the sum of the potential energy and the kinetic energy of the system in the second stage. Moreover, it should be noted that the initial velocity of the baseball is zero.
PE1+KE1=KE2+KE2
magh1+12mv12=magh2+12mv22
agh1=agh2+12mv22
v22=2agh1-h2
v22=2×9.798 ms2×300 m-0 m
v22=5878.8 m2s2
v2=76.67 ms
a=∆vt
t=∆va=76.67 ms9.798 ms2=7.825 seconds
Question #23
A. The Semimajor Axis
The semimajor axis of an ellipse refers to half of the major axis, which is the length of the ellipse. In orbital geometry, the semimajor axis (a) can be determined by dividing the sum of the radius of the apogee (Ra) and the radius of the perigee (Rp) by two (Sellers 2004, 135-136). In this problem, the radius of the perigee is equal to the sum of the radius of the Earth and the altitude of the perigee, and the radius of the apogee is equal to the sum of the radius of the Earth and the altitude of the apogee.
a=Ra+Rp2=(6378.137 km+2000 km)+(6378.137 km+375 km)2=7565.64 km
B. The Eccentricity
The eccentricity (e) of an ellipse refers to the ratio of the distance between the two foci of the ellipse (2c) and the major axis (2a). Note that the distance between the two foci (2c) is also equal to the difference between the radius of the apogee and the radius of the perigee (Sellers 2004, 135-136).
e=2c2a=Ra-RpRa+Rp
e=6378.137 km+2000 km-(6378.137 km+375 km)(6378.137 km+2000 km)+(6378.137 km+375 km)=0.1074
C. Altitude if the true anomaly is 175°
Note that the position of the spacecraft is defined by the eccentricity of the ellipse (as well as the length of the major axis) and the true anomaly, which is denoted as v (Sellers 2004, 136).
R=a1-e21+ecosv
R=7565.64 km×1-0.107421+0.1074×cos175°
R=8374.9 km
The altitude of the spacecraft can be calculated by subtracting the radius of the Earth from the computed radius.
Alt=R-REarth=8374.9 km-6378 km=1996.9 km
D. The flight-path angle when the true anomaly is 290°
The flight-path angle (ϕ) refers to the angle created by the local horizontal and the velocity vector. This value is positive when the spacecraft is travelling outbound (from perigee to apogee) and negative when travelling inbound (from apogee to perigee) (Sellers 2004, 136). In this case, we just have to determine whether the spacecraft is travelling inbound or outbound when the true anomaly is 290°.
At 290°, the aircraft must have completed reaching its orbital apogee, and is approaching the perigee. Therefore, the flight-path angle of the spacecraft when the true anomaly is 290° must be negative.
Question #25
A. Geocentric-equatorial coordinate system sketch
B. The specific angular momentum on the sketch
The specific angular momentum is calculated by determining the cross product of the position vector and the velocity vector.
h=R×V
h=7000 I+0 J+0 K×0 I-7.063 J+0 K
h=0 I+0 J-49441 K
h=-49441 km2s
C. Orientation of the orbit
Since the direction of the specific angular momentum is downward, it is safe to say that the direction of the orbit is from clockwise, and the orbital plane is perfectly horizontal.
D. Specific Mechanical Energy
Similar with the concept of specific angular momentum, the specific mechanical energy of a system resembles that of the mechanical energy, but the mass is canceled out. Therefore, the specific mechanical energy can be defined by the velocity of the spacecraft and its radius from the Earth’s center.
ε=V22-μR
ε=-7.063 kms22-3.986×105 km3s27000 km
ε=-32km2s2
E. Shape of Trajectory
Since the specific mechanical energy is negative, it is safe to say that the trajectory of the spacecraft is circular or elliptical.
Question #26
First, let us obtain the kinetic energy of the moving truck.
KE=mv22=2000 kg×65 mihr×1 hr3600 seconds×1.61 km1 mi22
KE=0.845 kgkm2s2
Now, let us consider the kinetic energy of the spacecraft. To determine the kinetic energy of the spacecraft, we have to solve for the velocity of the spacecraft, which can be calculated using the mean motion (n) and the circumference of its orbit.
n=μa3=3.986×105 km3s26378 km+759 km3=1.047×10-3 s-1
V=2πRn=2×π×6378 km+759 km×1.047×10-3 s-1
V=7.473 kms
We can now compute for the kinetic energy of the spacecraft.
KE=mv22=10,000 kg×7.473 kms22
KE=279,249 kgkm2s2
Reference
Sellers, Jerry Jon. Understanding Space: An Introduction to Astronautics. 2nd Edition. New York, NY: McGraw-Hill Companies, Inc., 2004.
