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. Furthermore, 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