This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Experiment 1, Physics 2BL Deduce the mean density of the Earth. 1. PHYSICS 1.1. Derivation of mean density in terms of gravity and radius, ( g,R E ) We will be determining the mean density of the Earth by calculating the radius of the earth, R E , and the accel- eration due to gravity at sea level, g . The force of gravity on a mass m at sea level is given by the following: F = GM E m R 2 E = mg Where G = 6 . 673 10- 11 m 3 kg s 2 . Next we solve for the mass of the Earth, M E in terms of the acceleration due to gravity at sea level, g : M E = gR 2 E G Which can be substituted into the expression for mean density. = M E 4 3 R 3 E = 3 g 4 GR E (1) 1.2. Measuring the radius of the Earth The diagram above requires a little imagination. The Sun is represented by the small circle that is making a 24 hour orbit (counterclockwise) around a large, non- rotating Earth. The tangent lines to the circle represent the event of a sunset. The first sunset happens at the bottom of the cliff, the second one at a height h above sea level. L is the distance to the horizon from the height of the cliff. Questions 1. In our experiment, who will see the sunset first? (The person on the top of the cliff or the person on the beach?) If this experiment were to be conducted on the East Coast, facing the Atlantic Ocean at sunrise, who would see the sunrise first? Draw a diagram similar to the one at the beginning of this section to explain the East Coast situation. We will now use geometry to solve for R E given quan- tities that we can measure. The Pythagorean Theorem gives us, L 2 + R 2 E = ( R E + h ) 2 which can be rearranged and simplified for h &lt;&lt; R E : L 2 = 2 R E h + h 2 2 R E h ; L p 2 R E h Now we set up a proportion to solve for the time in be- tween the two sunsets, t . This corresponds to the time it takes for the sun to go around the circumference of the Earth: L t = 2 R E 24 hrs We can now solve for the R E in terms of the height of the cliff, h , and the time between sunsets, t . We will substitute = 2 24 hrs . R E = 2 h 2 ( t ) 2 (2) Now lets re-derive this expression using the small angle aproximation approach. Please check Appendix 1 at the end of this lab guide to refresh your memory about small angles. This sort of argument is used very often in physics derivations. Looking at the figure we get the following relationship: cos( ) = R E R E + h Using Taylor Expansion for h/R E &lt;&lt; 1 on both sides gives: 2 1- 2 2 1- h R E Note that on the right hand side we used the expansion 1 1+ 1- where is a small value. Rearranging and substituting = t gives the same expression that we got in equation (2)....
View Full Document
- Spring '08