(a) Show that the rate of change of the free-fall acceleration with vertical position near the Earth's surface is dg 2GM 3--5 dr RE³ This rate of change with position is called a gradient. (b) Assuming his small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is |Agl = 2GMEh RE 3 (c) Evaluate this difference for h = 41.25 m, a typical height for a fifteen-story building. (Enter your answer in m/s².) m/s²

Using Figure 13.9, carefull sketch a free body diagram for the case of a simple pendulum hanging at latitude lambda, labeling all forces acting on the point mass,m. Set up the equations of motion for equilibrium, setting one coordinate in the direction of the centripetal accleration (toward P in the diagram), the other perpendicular to that. Show that the deflection angle , defined as the angle between the pendulum string and the radial direction toward the center of Earth, is given by the expression below. What is the deflection angle at latitude 45 degrees? Assume that Earth is a perfect sphere. tan(+)=gg2REtan , where is the angular velocity of Earth.

Check Your Understanding A top has a precession frequency of 5.0 rad/s on Earth. What is its precession frequency on the Moon?

On a planet whose radius is 1.2107m , the acceleration due to gravity is 18m/s2 . What is the mass of the planet?

Check Your Understanding Earth exerts a tidal force on the Moon. Is it greater than, the same as, or less than that of the Moon on Earth? Be careful in your response, as tidal forces arise from the difference in gravitational forces between one side and the other. Look at the calculations we performed for the tidal force on Earth and consider the values that would change significantly for the Moon. The diameter of the Moon is one-fourth that of Earth. Tidal forces on the Moon are not easy to detect, since there is no liquid on the surface.

Calculate the effective gravitational field vector g at Earths surface at the poles and the equator. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. How well does the result agree with the difference calculated with the result g = 9.780356[1 + 0.0052885 sin 2 0.0000059 sin2(2)]m/s2 where is the latitude?

Show that for eccentricity equal to one in Equation 13.10 for conic sections, the path is a parabola. Do this by substituting Cartersian coordinates, x and y, for the polar coordinates, r and , and showing that it has the general form for a parabola, x=ay2+by+c .

Check Your Understanding Galaxies are not single objects. How does the gravitiational force of one galaxy exerted on the “closer” stars of the other galaxy compare to those farther away? What effect would this have on the shape of the galaxies themselves?

Let gM represent the difference in the gravitational fields produced by the Moon at the points on the Earths surface nearest to and farthest from the Moon. Find the fraction gM/g, where g is the Earths gravitational field. (This difference is responsible for the occurrence of the lunar tides on the Earth.)

Using the technique shown in Satellite Orbits and Energy, show that two masses m1 and m2 in circular orbits about their common center of mass, will have total energy E=K+E=K1+K2Gm1m2r=Gm1m22r . We have shown the kinetic energy of both masses explicitly. (Hint: The masses orbit at radii r1 and r2 , respectively, where r=r1+r2 . Be sure not to confuse the radius needed for centripetal acceleration with that for the gravitational force.)