10.Determine the force of gravitational attraction between the earth (m=5.98x10^24 kg) and a 70 kg physics student is standing at sea level, a distance of 6.38x10*6 m from earth's center. O 680 N O 686 N 658 N O 706 N O O

1. You and a friend set up an umbrella and chairs at Torrey Pines State Beach. Your friend goes boogie boarding while you read a book. Half an hour later you are surprised to see that your friend has moved south, parallel to the beach. They are still boogie boarding the same distance from shore, in the surf zone. Why did your friend drift parallel to the beach? 2. The Sun is much more massive than the Moon, but the Moon has a bigger influence on Earth's tides than the Sun. Why is this? 3.  Even when there is no wind at the beach, we still see waves. What is the source of these waves? Do not include tides, mass movement, or earthquakes in your answer. 4. What causes ocean surface water to sink to the bottom and become deep water? Where does this happen?  5. How do you escape a Rip Current?

Question 1 The Oort Cloud extends out to (possibly) one light- year from the sun. Objects in the Oort Cloud are still gravitationally bound to the sun. Suppose one such iceball orbits the sun in a circle. I'm going to alter some numbers, such as the mass of the sun and even G. The gravitational force is directed toward the sun and has the following magnitude: GMm p2 Calculate the force on the object if these are the numbers: G = 5.1*10-11 N*m?/kg² • M = 2.4*1030 kg • m = 1.0*10® kg r = 7*1015 m Calculate your answer in microNewtons (10-6 N).

A satellite in geostationary orbit (also called synchronous orbit) appears to remain stationary in the sky as seen from any particular location on the planet.  a.) In the future, there will be need for satellites in synchronous orbit around Mars to aid colonies. At what altitude would such a satellite need to be above the surface of Mars?Assume that the mass of Mars is 6.39 × 10^23 kg, the length of the Martian solar day (i.e., sol) is 24h 39m 35s, the length of the sidereal day is 24h 37m 22s, and the equatorial radius is 3396 km. (Hint: if you haven’t had a physics class before, you can find this by using the fact that the acceleration of an object in circular motion either as v2/r, where v and r are the velocity and radius of the orbit, or as 4Pi 2r/T2 , where T is the period. Use this second equation and Mathematical Insight 4.5 on p. 131 to find r for T=1 day. Make sure to use values for Mars nstead of Earth, as necessary. Alternatively, you can calculate the answer using Newton’s…