Sandra Miller and Stephanie Smith Lamar High School Arlington, TX

This problem is designed to occur during a Geometry unit on circles.  A line tangent to a circle forms a right angle with a radius drawn at the point of tangency.

 r – radius of the planet/moon  h – height of the observer (eyes)  d – distance to the horizon r r h d

 r – radius of the planet/moon  h – height of the observer (eyes)  d – distance to the horizon r r h d

ObjectRadiusHorizon Earth3959 mi.3 mi. Moon1080 mi. Mars2106 mi. Jupiter43,441 mi.

ObjectRadiusHorizon Earth3959 mi.3 mi. Moon1080 mi.1.6 mi. Mars2106 mi.2.2 mi. Jupiter43,441 mi.9.9 mi.

 This problem set is geared toward a Pre-AP Algebra I class or an Algebra II class.  By working through this packet, a student will practice  Simplifying literal equations  Creating formulas  Unit conversions  Using formulas to solve problems

Sir Isaac Newton developed three equations that we will use to develop some interesting information about the solar system. When a force F acts on a body of mass m, it produces in it an acceleration a equal to the force divided by the mass. The centripetal acceleration a of any body moving in a circular orbit is equal to the square of its velocity v divided by the radius r of the orbit. The grativational force F between two objects is proportional to the product of their two masses, divided by the distance between them.

 If we substitute the formula for centripetal acceleration into the F = ma equation, we have an equation for the orbital force:  The gravitational force that the object being orbited exerts on its satellite is

 Objects that are in orbit stay in orbit because the force required to keep them there is equal to the gravitational force that the object being orbited exerts on its satellite.  If we set our two equations equal to each other and solve for v, we end up with a formula that will give us the orbital speed of the satellite.

 Simplify the equation and solve for v:

 Because the mass of the satellite m cancelled out of the equation, if we know the orbital velocity and the radius of the orbit, we can find the mass of the object being orbited.

 Rewrite the velocity equation and solve for M:

 Example: Use the Moon to calculate the mass of the Earth.  Orbital radius:  Period: T = 27.3 days  Orbital velocity:

 Example: Use the Moon to calculate the mass of the Earth.

 To calculate escape velocity, we set the equation for kinetic energy to the equation for gravitational force and solve for v: Kinetic energy > Force × distance

Calculate Earth’s escape velocity in km/s.  Earth’s mass: 6.02 × kg  Earth’s radius:6.38 × 10 6 m

 Now that we’ve worked through the different equations, we can calculate the mass and escape velocity of Mars as well as the mass of the Sun.

One of my favorite sites for possible astronomy-related math problems has been Space Math at Unfortunately, because of cutbacks in NASA’s education budget, it will not be updated as frequently.

Original (Standard) Problem Invert the problem Ask for prediction Break into multiple parts Ask for multiple representation Ask questions that require qualitative reasoning Automaticity practice Ask for generalization Examples or counter- examples Ask for an explanation: oral or written James Epperson, Ph.D.

 The powerpoint and the worksheets will be posted on my blog at tothemathlimit.wordpress.com.