Motor Daddy Posted June 18, 2008 Share Posted June 18, 2008 Let's just ASSUME the Earth has the mass of 5,974,200,000,000,000,000,000,000 kg. We are going to compare impact times of two different objects when dropped from an exact height of 16.087 feet. Object A has a mass of 1 kg Object B has a mass of 10 kg Using the formula A=(L-S)/R2 Object A has a "A value" of 371,368,185,491,390,563,809,286.93976503 Object B has a "A value" of 371,368,185,491,390,563,809,286.38030708 A previous test was done with object A. It was determined that object A took exactly 1 second to impact the ground when dropped from a height of 16.087 feet, which is an acceleration of 32.174 ft/sec^2. That means an “A value” of 1, has an acceleration of 32.174/371,368,185,491,390,563,809,286.93976503= .000000000000000000000086636392822469954136118658002141 ft/sec^2 If you multiply that by Object B’s “A value”, you find an acceleration of 32.173999999999999999999951530582 ft/sec^2 for object B. Now, let’s look at the time it takes for each object to hit the ground, when dropped from the 16.087 feet. Object A- 1.0000000000000000000000000000000 seconds Object B- 1.0000000000000000000000007532389 seconds Link to comment Share on other sites More sharing options...
Klaynos Posted June 18, 2008 Share Posted June 18, 2008 What do all your symbols mean? And you should use: [math]g=-\frac{GM} {r^2}[/math] To be perfectly specific you need to do accelerations for the earth towards the object as well... Link to comment Share on other sites More sharing options...
ecoli Posted June 18, 2008 Share Posted June 18, 2008 the acceleration of the two objects shouldn't be different, make sure you're using the right equations. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now