美国麻省理工大学：《动力工程学》（英文版）Lecture D15-Gravitational Attraction. The Earth as a Non-Inertial Reference Frame

16.07 Dynamics Fall 2004 Version 1.3 Lecture d15- Gravitational Attraction. The earth as a Non-Inertial reference frame Gravitational attraction The Law of Universal Attraction was already introduced in lecture Dl. The law postulates that the force of attraction between any two particles, of masses M and m, respectively, has a magnitude, F, given by Mm where r is the distance between the two particles, and g is the universal constant of gravitation. The value of G is empirically determined to be 6673(10(-11)m /(kg. s)2. The direction of the force is parallel to the line connecting the two particles Recall, from lecture D8, that the gravitational force is a conservative force that can be derived from a potential. The potential for the gravitation force is given by F=-Vv The law of gravitation stated above is strictly valid for point masses. One would expect that when the when e sizes of the masses are comparable to the distance between the masses, one would observe deviations to the above law. In such cases, the forces due to gravitational attraction would depend on the spatial distribution of the mass Consider the case in which the mass m has a small size and can be regarded as a point mass, whereas the size of mass M is large compared to the distance between the two masses d

J. Peraire 16.07 Dynamics Fall 2004 Version 1.3 Lecture D15 - Gravitational Attraction. The Earth as a Non-Inertial Reference Frame Gravitational attraction The Law of Universal Attraction was already introduced in lecture D1. The law postulates that the force of attraction between any two particles, of masses M and m, respectively, has a magnitude, F, given by F = G Mm r 2 (1) where r is the distance between the two particles, and G is the universal constant of gravitation. The value of G is empirically determined to be 6.673(10( − 11))m3/(kg.s) 2 . The direction of the force is parallel to the line connecting the two particles. Recall, from lecture D8, that the gravitational force is a conservative force that can be derived from a potential. The potential for the gravitation force is given by V = −G Mm r , and F = −∇V . The law of gravitation stated above is strictly valid for point masses. One would expect that when the when the sizes of the masses are comparable to the distance between the masses, one would observe deviations to the above law. In such cases, the forces due to gravitational attraction would depend on the spatial distribution of the mass. Consider the case in which the mass m has a small size and can be regarded as a point mass, whereas the size of mass M is large compared to the distance between the two masses. 1

In this case, the potential energy is given by V=-gm dM That is, the total potential energy is the sum of the potential energies due to small elemental masses, dM The integration must be carried out over the entire mass M, where r is the distance between m and the elemental mass dm being considered It turns out that if the mass M, is distributed uniformly over a spherical shell of radius R, then it can be hown, by carrying out the above integral, th le potential when m is inside the shell is constant and equal to Mm In this case. we have F=-VV=0 The potential when m is outside the shell is given by Mm where r is the distance from m to the center of the shell. In this case, the potential, and, consequently, the force, is identical to that of a point mass M located at the center of the spherical shell Therefore, when the mass M is a solid sphere, the gravitational attraction on a mass m, outside M, is stil given by(1), with r being measured from the sphere center F If, on the other hand, the mass m is inside m, then the attraction force on m due to m, is given by Mm

In this case, the potential energy is given by V = −Gm Z M dM r . That is, the total potential energy is the sum of the potential energies due to small elemental masses, dM. The integration must be carried out over the entire mass M, where r is the distance between m and the elemental mass dM being considered. It turns out that if the mass M, is distributed uniformly over a spherical shell of radius R, then it can be shown, by carrying out the above integral, that: • The potential when m is inside the shell is constant and equal to V = −G Mm R . In this case, we have F = −∇V = 0 • The potential when m is outside the shell is given by V = −G Mm r , where r is the distance from m to the center of the shell. In this case, the potential, and, consequently, the force, is identical to that of a point mass M located at the center of the spherical shell. Therefore, when the mass M is a solid sphere, the gravitational attraction on a mass m, outside M, is still given by (1), with r being measured from the sphere center. If, on the other hand, the mass m is inside M, then the attraction force on m due to M, is given by F = G M′m r 2 = G Mm R2 ( r R ) . 2

Here, M'=M(r/R)is the mass corresponding to a hypothetical sphere of radius r with the same uniform mass density as the original sphere of radius R. In other words, the force of attraction when m is inside M is equal to that of a reduced sphere of radius r instead of R. Thus, we see that the spherical shell outside m, has no effect on the gravitational attraction force on m ght Weigl The gravitational attraction from the Earth to any particle located near the surface of the Earth is called the weight. Thus, the weight, w, of a particle of mass m, is given approximately by W≈-G gomer = go Here, Me and R are the mass and radius of the Earth, and go =-(GMeR er is called the gravitational acceleration vector It turns out that the Earth is not quite spherical, and so the weight does not exactly obey the inverse-squared law. The magnitude of the gravitational acceleration, go, at the poles and at the equator is slightly different In addition, the earth is also rotating. This introduces an inertial centrifugal force which has the effect of reducing the vertical component of the weight Note Gravity variations due to Earth rotation Here, we consider the influence of Earth's rotation on the gravity measured by an observer rotating with the Earth. The starting point will be our general expression for relative motion F-maB-2mg×(vA/B)xy2-m9xr4/B-m9x(gxrA/B)=m(aA/B)x'y2·(2) We consider two reference frames. A fixed frame ayz, and a frame r'y'z' that rotates with the earth. Both the inertial observer, O, and the rotating observer, B, are situated at the center of the Earth, and are observing a mass m situated at point A on the Earths surface (Sxra/B) O

Here, M′ = M(r/R) 3 is the mass corresponding to a hypothetical sphere of radius r with the same uniform mass density as the original sphere of radius R. In other words, the force of attraction when m is inside M is equal to that of a reduced sphere of radius r instead of R. Thus, we see that the spherical shell outside m, has no effect on the gravitational attraction force on m. Weight The gravitational attraction from the Earth to any particle located near the surface of the Earth is called the weight. Thus, the weight, W, of a particle of mass m, is given approximately by W ≈ −G Mem R2 e er = −g0mer = mg0 . Here, Me and R are the mass and radius of the Earth, and g0 = −(GMe/R2 e )er is called the gravitational acceleration vector. It turns out that the Earth is not quite spherical, and so the weight does not exactly obey the inverse-squared law. The magnitude of the gravitational acceleration, g0, at the poles and at the equator is slightly different. In addition, the Earth is also rotating. This introduces an inertial centrifugal force which has the effect of reducing the vertical component of the weight. Note Gravity variations due to Earth rotation Here, we consider the influence of Earth’s rotation on the gravity measured by an observer rotating with the Earth. The starting point will be our general expression for relative motion, F − m aB − 2m Ω × (vA/B)x′y′z ′ − m Ω˙ × rA/B − m Ω × (Ω × rA/B) = m (aA/B)x′y′z ′ . (2) We consider two reference frames. A fixed frame xyz, and a frame x ′ y ′ z ′ that rotates with the Earth. Both the inertial observer, O, and the rotating observer, B, are situated at the center of the Earth, and are observing a mass m situated at point A on the Earth’s surface. 3

The forces on the mass will be the gravitational force, mgo, and the reaction force, R, which is needed te keep the mass at rest relative to the Earth's surface (if the mass m is placed on a scale, R would be the force that the scale exerts on the mass). Thus, F=R+mgo. Since the mass m is assumed to be at rest =0,and,O≡B, we have, R+mgo-mnX(SXTA/B)=0, or, R=-m[ X(SX TA/Bl Thus, an observer at rest on the surface of the Earth will observe a gravitational acceleration given by g=go-QX( XTA/B). The term -nx(QXTA/B) has a magnitude S-d=S-Re cos L, and is directe normal and away from the axis of rotation An alternative choice of reference frames which is sometimes more convenient when working with the Earth s a rotating reference frame is illustrated in the figure below a y The fixed ayz axes are the same as before, but now the rotating observer B is situated on the surface of the Earth. A convenient set of rotating axes is that given by Nort-West-South directions ry'z'. If we assume that the mass m is located at B, then we have a= B, and the above expression(2)reduces to R+mgo -maB=0 mIgo -aB=-mg It is straightforward to verify that aB=Sx(Sx rB), which gives the same expression for R, as expected If we call g the gravity acceleration vector, which combines the fact that the Earth is not spherical and that it is rotating. the g is gl g≈9780327(1+0.005279sin2L+0.00046in4L) where L is the latitude of the point considered and g is given in m/s. The coefficient 0.005279 has two components: 0.00344, due to Earth's rotation, and the rest is due to Earth,s oblateness(or lack of sphericity)

The forces on the mass will be the gravitational force, mg0, and the reaction force, R, which is needed to keep the mass at rest relative to the Earth’s surface (if the mass m is placed on a scale, R would be the force that the scale exerts on the mass). Thus, F = R + mg0. Since the mass m is assumed to be at rest, Ω˙ = 0, and, O ≡ B, we have, R + mg0 − m Ω × (Ω × rA/B) = 0 , or, R = −m[g0 − Ω × (Ω × rA/B] = −mg Thus, an observer at rest on the surface of the Earth will observe a gravitational acceleration given by g = g0 − Ω × (Ω × rA/B). The term −Ω × (Ω × rA/B) has a magnitude Ω2d = Ω2Re cosL, and is directed normal and away from the axis of rotation. An alternative choice of reference frames which is sometimes more convenient when working with the Earth as a rotating reference frame is illustrated in the figure below. The fixed xyz axes are the same as before, but now the rotating observer B is situated on the surface of the Earth. A convenient set of rotating axes is that given by Nort-West-South directions x ′y ′ z ′ . If we assume that the mass m is located at B, then we have A ≡ B, and the above expression (2) reduces to R + mg0 − m aB = 0 , or, R = −m[g0 − aB] = −mg . It is straightforward to verify that aB = Ω × (Ω × rB), which gives the same expression for R, as expected. If we call g the gravity acceleration vector, which combines the fact that the Earth is not spherical and that it is rotating, the magnitude of g is given by g ≈ 9.780327(1 + 0.005279 sin2 L + 0.000024 sin4 L), where L is the latitude of the point considered and g is given in m/s2 . The coefficient 0.005279 has two components: 0.00344, due to Earth’s rotation, and the rest is due to Earth’s oblateness (or lack of sphericity). 4

The higher order term is also due to oblateness. The above expression is known as the international gravity 984 Relative to non-rotating earth 983 982 981 98 Relative to rotating earth 979 978 102030405060708090 uator Latitude(degrees) (Poles We note that the gravitational acceleration at the poles is about 0. 5% larger than at the equator. Further more, the deviations due to the earth's rotation are about three times larger than the deviations due to the Earth’ oblateness Angular deviation of g Here, we consider a spherical Earth, and we want to determine the effect of Earth's rotation on the direction In the previous note, we established that an observer rotating with the Earth will observe a gravity vector g=g where go is the geocentric gravity, and g is the modified gravity R cos L Re

The higher order term is also due to oblateness. The above expression is known as the international gravity formula and is depicted below. 0 10 20 30 40 50 60 70 80 90 9.77 9.78 9.79 9.8 9.81 9.82 9.83 9.84 (Equator) (Poles) Latitude (degrees) m/s 2 Relative to non-rotating earth Relative to rotating earth We note that the gravitational acceleration at the poles is about 0.5% larger than at the equator. Further￾more, the deviations due to the Earth’s rotation are about three times larger than the deviations due to the Earth’ oblateness. Note Angular deviation of g Here, we consider a spherical Earth, and we want to determine the effect of Earth’s rotation on the direction g. In the previous note, we established that an observer rotating with the Earth will observe a gravity vector given by g = g0 − Ω × (Ω × r) , where g0 is the geocentric gravity, and g is the modified gravity. 5

From the triangle formed by go, g, and (22Rcos L, we have g sin &=02 Rcos Lsin L=(32R/2)sin2L.We expect 8 to be small, and, therefore sin d A 0, and g A go. Thus ArE which is maximum when L =+45. In this case, we have Q2=7. 29(10-)rad /s, Re=6370 km, and 6max=17(10-3)rad≈0.1° We now consider a couple of three dimensional examples. In the first example, the motion is known, and we are asked to determine the forces required to obtain that motion. In the second example, the motion is unknown and the trajectory needs to be obtained by integrating the equation of motion. Example Aircraft flying at constant velocity We now consider an aircraft A, flying with constant velocity v= UNeN Uwew uvey relative to the surface of the earth. We assume our inertial observer to be at the center of the earth. and our accelerated observer to be at the aircraft(e. g. A= B) y The angular velocity of the Earth, n, can be expressed as n=S cos LeN +Ssin Leu. The aerodynam force, R, that an aircraft is required to generate in order to maintain its course is R+mgo-maB-2mn XUNWU =0 Since aB=92×(!xTB),andg=go-×(g×rB)≈-geU, we have, R=-2mQww sin LeN +2mQ(UN sin L-vU cos L)ew+(mg+ 2mnuw cos L)er For instance, for an aircraft to fly horizontally (i.e. vu =0), it will require a horizontal force, RH 2mQsin L(-WW eN +UN ew). We see that for L>0(northern hemisphere), this force is always to the"left of the aircraft, and needs to be generated aerodynamically in order to maintain a straight path. The reverse true in the southern hemisphere

From the triangle formed by g0, g, and Ω2R cosL, we have g sin δ = Ω2R cosL sinL = (Ω2R/2) sin 2L. We expect δ to be small, and, therefore, sin δ ≈ δ, and g ≈ g0. Thus, δ ≈ Ω 2Re 2g sin 2L , which is maximum when L = ±45o . In this case, we have Ω = 7.29(10−5 ) rad/s, Re = 6370 km, and δmax = 1.7(10−3 ) rad ≈ 0.1 o . We now consider a couple of three dimensional examples. In the first example, the motion is known, and we are asked to determine the forces required to obtain that motion. In the second example, the motion is unknown and the trajectory needs to be obtained by integrating the equation of motion. Example Aircraft flying at constant velocity We now consider an aircraft A, flying with constant velocity v = vN eN + vW eW + vU eU relative to the surface of the Earth. We assume our inertial observer to be at the center of the Earth, and our accelerated observer to be at the aircraft (e.g. A ≡ B). The angular velocity of the Earth, Ω, can be expressed as Ω = Ω cosL eN + Ω sinL eU . The aerodynamical force, R, that an aircraft is required to generate in order to maintain its course is R + mg0 − maB − 2mΩ × (v)NW U = 0 . Since aB = Ω × (Ω × rB), and g = g0 − Ω × (Ω × rB) ≈ −geU , we have, R = −2mΩvW sinL eN + 2mΩ(vN sinL − vU cosL) eW + (mg + 2mΩvW cosL) eU . For instance, for an aircraft to fly horizontally (i.e. vU = 0), it will require a horizontal force, RH = 2mΩ sinL(−vW eN +vN eW ). We see that for L > 0 (northern hemisphere), this force is always to the “left” of the aircraft, and needs to be generated aerodynamically in order to maintain a straight path. The reverse is true in the southern hemisphere. 6

H H For a 200 Ton(m=2(10%)kg) aircraft at 300 m/s, at a latitude of L=42 north, the magnitude of this force is RH=5840 N=1320 Ib Note that if this force is not provided, the aircraft will turn to its right We also see that for the same aircraft flying east, there is an extra upwards lift of magnitude 2mQww cos L For our aircraft, that amounts to 6490 N= 1460 lb, 0. 3% of its weight, or about 7 extra passengers Falling ob Consider an object being released from a point P situated at a height of 200 m. Calculate the distance between the point of impact and the point at which the plumb line going though P intersects the ground Neglect air resistance, and assume L= 45 We consider the rotating right handed set of axes NwU, and write v= uneN wwew uvev and Q=Scos LeN +Qsin Ley. The Coriolis acceleration is thus 222 x(U)NWU=2 9 cosL 0 QsinL 2Q2 sin Lww eN 2Q(UN sin L-UU cos L)ew 2S2 cos Lww eu Since we expect UU >UN, ww, we retain only the effect of wU. Thus, acor -292 cos Luv ew. The equation of motion for a falling object will therefore be -mg ev-macor=m(a)NwU Here, g includes the centrifugal effects, and (a)NwU is the acceleration experienced by a non-inertial observer on the earth's surface In the ev direction, we have au =-, vu = -gt, and Iu=200-(gt2)/2. The time required for the object to reach the ground will be obtained for Tu=0, which gives t= 6.4 s. In the ew direction, we have 20 cos L 2Q2g cos Lt, Uw=-ng cos Lt, and Iw =-ng cos L(t/ 3). For t=6.4 s, this gives xW=-0.044m≈-4cm

For a 200 Ton (m = 2(105 )kg) aircraft at 300 m/s, at a latitude of L = 42o north, the magnitude of this force is RH = 5840 N = 1320 lb. Note that if this force is not provided, the aircraft will turn to its right. We also see that for the same aircraft flying east, there is an extra upwards lift of magnitude 2mΩvW cosL. For our aircraft, that amounts to 6490 N = 1460 lb, 0.3% of its weight, or about 7 extra passengers. Example Falling object Consider an object being released from a point P situated at a height of 200 m. Calculate the distance between the point of impact and the point at which the plumb line going though P intersects the ground. Neglect air resistance, and assume L = 45o . We consider the rotating right handed set of axes NW U, and write v = vN eN + vW eW + vU eU and Ω = Ω cosL eN + Ω sinL eU . The Coriolis acceleration is thus acor = 2Ω × (v)NW U = 2          eN eW eU Ω cosL 0 Ω sinL vN vW vU          = −2Ω sinLvW eN + 2Ω(vN sinL − vU cosL) eW + 2Ω cosLvW eU Since we expect vU ≫ vN , vW , we retain only the effect of vU . Thus, acor ≈ −2Ω cosLvU eW . The equation of motion for a falling object will therefore be, −mg eU − macor = m(a)NW U . Here, g includes the centrifugal effects, and (a)NW U is the acceleration experienced by a non-inertial observer on the Earth’s surface. In the eU direction, we have aU = −g, vU = −gt, and xU = 200 − (gt2 )/2. The time required for the object to reach the ground will be obtained for xU = 0, which gives t = 6.4 s. In the eW direction, we have aW = 2Ω cosLvU = −2Ωg cosLt, vW = −Ωg cosLt2 , and xW = −Ωg cosL(t 3/3). For t = 6.4 s, this gives xW = −0.044 m ≈ −4 cm. 7

Example(Adapted from MMS (Optional) Cyclonic Air Motion Suppose that there is an area of low pressure in the Northern Hemisphere, so that the pressure force per unit mass on an air element is -Vp/p and is radially inwards. One would think that the air should rush in radially under this force to“ fill in the hole” eu Instead, the wind may be such that air moves in circular paths around the depression. The radial acceleration (ar)r'y?=-r02=-v2/r. The real force acting radially per unit mass is, as noted, -(1/p)dp/dr, nd, in addition, we would have to include the inertial forces due to Earth's rotation, namely, a Coriolis force x(o)r'y'2' per unit mass. The combined effect of gravity and Earth's centrifugal force acts in the direction of the local vertical. Therefore, the equation of motion in the radial direction is ug 1 dp +290 dr Here, Q u =n ey is the component of the angular velocity in the vertical direction. If we compare the magnitude of the acceleration term with that due to Coriolis effect, 29u ve see that for a given Qu, Coriolis' effect becomes important for large values of r/vg. Therefore, we consider two limits Large values of r/ve. This leads to the so called Geostrophic Winds. In this case, the acceleration term nall and the appl g equation becomes 0= I dp +29U Consider for instance ve=10 m/s and r in the range of 100 to 400 km. At a latitude of 42, we have QU A 4.9 x 10- rad/s, and dp/dr=-2p Quvg A 0.0012 N/m, or 1.2 mb per 100 km, a moderate

Example (Adapted from MMS) (Optional) Cyclonic Air Motion Suppose that there is an area of low pressure in the Northern Hemisphere, so that the pressure force per unit mass on an air element is −∇p/ρ and is radially inwards. One would think that the air should rush in radially under this force to “fill in the hole”. Instead, the wind may be such that air moves in circular paths around the depression. The radial acceleration is then, (ar)x′y′z ′ = −r ˙θ 2 = −v 2 θ /r. The real force acting radially per unit mass is, as noted, −(1/ρ)dp/dr, and, in addition, we would have to include the inertial forces due to Earth’s rotation, namely, a Coriolis force −2Ω × (v)x′y′z ′ per unit mass. The combined effect of gravity and Earth’s centrifugal force acts in the direction of the local vertical. Therefore, the equation of motion in the radial direction is − v 2 θ r = − 1 ρ dp dr + 2ΩU vθ . Here, ΩU = Ω · eU is the component of the angular velocity in the vertical direction. If we compare the magnitude of the acceleration term with that due to Coriolis’ effect, 2ΩU vθ v 2 θ /r = 2ΩU r vθ , we see that for a given ΩU , Coriolis’ effect becomes important for large values of r/vθ. Therefore, we consider two limits: • Large values of r/vθ. This leads to the so called Geostrophic Winds. In this case, the acceleration term is small and the approximate governing equation becomes, 0 = − 1 ρ dp dr + 2ΩU vθ . Consider for instance vθ = 10 m/s and r in the range of 100 to 400 km. At a latitude of 42o , we have ΩU ≈ 4.9 × 10−5 rad/s, and dp/dr = −2ρΩU vθ ≈ 0.0012 N/m3 , or 1.2 mb per 100 km, a moderate 8

pressure gradient. These winds are responsible for most regional weather patterns; they circulate ounterclockwise in the Northern Hemisphere(and clockwise in the Southern Hemisphere Small values of r/ve. This limit includes the tornadoes. Pressure defects of the order of 0.1 atm a 0. 1 x105 b can occur in a tornado over scales of the order of 10 m. In this case. Coriolis'effect become unimportant and the governing equation reduces to dp Typical values for the velocity are ≈100m/s References 1 M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96

pressure gradient. These winds are responsible for most regional weather patterns; they circulate counterclockwise in the Northern Hemisphere (and clockwise in the Southern Hemisphere). • Small values of r/vθ. This limit includes the tornadoes. Pressure defects of the order of 0.1 atm ≈ 0.1 × 105 b can occur in a tornado over scales of the order of 10 m. In this case, Coriolis’ effects become unimportant and the governing equation reduces to − v 2 θ r = − 1 ρ dp dr . Typical values for the velocity are, vθ = s r ρ dp dr ≈ 100m/s . References [1] M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96. 9