Announcements Exam 1 is next week. Will be similar to homework problems. Bring a scientific calculator. Equations Sheet will be provided (same one you have been using). Posted lectures have worked out solutions to the example problems. Solutions to all but the HW assigned today will be posted by Friday afternoon. Solutions to today’s HW will be posted Monday Homework Set 4: Chapter 4 # 50 & 52 + Supplemental Problems

Gravity and Tidal Stress

In addition to his three laws of motion, Newton formulated the Universal Law of Gravity G is the universal gravitational constant (6.67 x ), m 1 and m 2 are the masses of the two objects, r 1-2 is the distance between them and is the direction of the line connecting the two masses.

Easy example Using the information in the Appendix, determine the gravitational force of the Earth on the Moon. Which way does it act? What is the gravitational force of the Moon on the Earth? Which way does it act?

Solution M E = 5.97 x kgM M = 7.35 x kg r E-M = x 10 5 km Gravity is always an attractive force so the Earth pulls on the Moon: the direction is from the Moon to the Earth. The force of the Moon on the Earth is exactly the same but in the opposite direction (Newton’s 3 rd Law) 1.98 x N directed from the Earth to the Moon

What would the gravitational force be inside the Earth? Newton showed that, as long as the body is spherically symmetrical, the only thing that matters is the mass of the sphere underneath you. The spherical shell above you does not contribute to the gravitational force. M’ is the mass of the sphere with radius (R-d) and d is the distance below the surface

Gravitational acceleration decreases as you move inward This assumes a uniform density. For a real planet, the density increases as you go down so the acceleration increases at first then starts to decrease.

Example Assuming a uniform density, what is the acceleration due to gravity at one fifth of the way to the center of the Earth?

Example Solution First, find out how much mass is below: Use (R-d) = 4 / 5 R earth = 0.8 x 6.378x10 6 m = x 10 6 m and  = 5,515 kg / m 3

Example Solution continued Now find “g” from the gravitation formula Use M’= 3.069x10 24 kg and (R-d) = 5.102x10 6 m

Tides are the result of a difference in gravity

Mathematical treatment of tidal force Start with the gravitational force between two point masses a distance r apart Now make M a sphere with radius  r so that a point on the surface of M is R ±  r from m (ignore the directions for now) thus

Mathematical treatment of tidal force Pull out an R 2 term in the denominator Now expand this in a Maclaurin series expansion The first term is just the normal gravitational force between the two masses. The second term (and other terms) is the tidal force

So the tidal force goes as one over R 3 Note that  r is the radius of the body the tides are acting on and R is the center-to- center distance between the two bodies.

Example Calculate the tidal force of the Earth acting on a 1kg mass on the surface of the Moon For this problem m is the mass of the Earth ( x kg), M is the one kilogram mass,  r is the radius of the Moon ( x 10 6 m) and R is the Earth-Moon distance (3.844 x 10 8 m)

The Roche Limit When the tidal force on a body exceeds the binding force between the particles the body is composed of, it is at the Roche Limit. If the binding force holding an object together is simply its self gravity, the Roche limit is R is the radius of the primary body,  M is the density of the primary and  m is the density of the moon

Examples Determine the Roche Limit for Earth. The ISS orbits at a distance of 6,778 km from the center of the Earth. Is it inside the Roche limit? If so, why doesn’t it come apart?

Solution From Appendix 4: average density of Earth is 5513 kg / m 3 and radius is 6378 km. From a web search: ISS volume = 916 m 3 and mass = 419,455 kg Average density of ISS:

Example Solution continued Plug densities and radius into Roche formula So, the ISS orbits well within the Roche Limit of Earth. Why doesn’t it come apart?