Question

Newton’s Cannon Imagine you fired a cannonball such that the range of that ball was slightly longer than the radius of the Earth. As that ball fell back down, it would miss the ground and continue to fall. If we can ignore any air resistance or collisions with other objects, then we can say the ball would continue to fall forever, just missing the ground, and thus be in orbit. Assume all orbits here are perfectly circular and pretend that the Earth is perfectly spherical with a radius of 6370 km and a mass of 5.97x10^24 kg. (Use g=9.8 m/s² and G= 6.67x10^-11 N m²/kg²)

1. Setup a cannon on the surface of the Earth to launch a 10 kg cannonball at 45°. Draw a picture of the setup, including showing the curvature of the Earth (need not be to scale). Using the range equation, how fast would this cannonball have to go to just miss the Earth as it fell back down?

2. As the cannonball falls, it creates a circular orbit. Draw a picture of this orbit, labeling all useful/relevant quantities and a FBD of the cannonball. Using your tools from uniform circular motion, calculate how fast the cannonball must travel to maintain this path. Does it agree with part 1?

3. Communication satellites orbit the Earth at a height of 36,000 km above the surface. Assume we put these satellites into orbit using cannons, how fast would you have to launch a 4000 kg satellite from the surface of Earth to get it into orbit that high?

4. Using our tools from uniform circular motion, recalculate the required velocity of the orbital satellite, but this time use Newton’s Universal Law of Gravitation to calculate the centripetal force keeping it in orbit. Does your number agree with 3? Why or why not?

5. How far out would a satellite orbit if it had the velocity calculated from part 3? Compare that number to the radius of the Earth.

6. The center to center distance between the Earth and the moon is about 385,000 km apart. If the moon orbits the Earth every 27.4 days, how long does it take the satellite from part 3 to orbit the Earth? (Keppler’s 3rd Law)

Questions 5 and 6 are the only ones needed

Answer 1

(5) It can be calculated using concept of orbital velocity ,

GM V =

(5) For Part (3) final Eneryfsfel Satellite nitial Energ - Totl nergfinal GMMS m-GMe -GMeMs 2K 6-65Re Re 2 GMe 2 1.248 GMe Re XO 924 as Collated nart 31 Noro fr rt (5) fand Sarle llite is as ebeity in fart (3) 00 Orhital hane Amet we GMe orbital vebety V. GM GMe 1842 GMe Re Re 3446.96 K 1248 0 This Tadius TRSatellik will Cla wite eanth

(6) It is calculated by using Keppler's law

(6) rom kep pler's 3'd lan, use have 2 So, 3 38S 36 oD 3 2 /36 =) T 274 385 20 MAt 385 2न-५ = 0.8425 a - 20.22 hrs Hert, Time feried T= gafellite 11 2/E: