Distance travelled of an object falling from non-negligible height

  • Thread starter kaikalii
  • Start date
  • #1
17
0
In a free-fall problem, 9.8 m/s^2 works fine as long as you stay close to the earth. However, if you are falling from very high up, the change in acceleration due to gravity is no longer negligible.

Let's say there's an atmosphereless planet of mass M, and another object of negligible mass is falling toward it after being dropped from a point a distance r from the center of the planet. How could you derive an equation to find the distance fallen d as a function of time t and initial height r? Just so you are aware, I have a very limited knowledge of calculus (how to find derivatives and integrals), so please answer in terms I can understand.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
You will need to learn more calculus in order to understand this sort of physics.
If you just do F=ma as usual you get:
$$\frac{GM}{r^2(t)} = \frac{d^2}{dt^2}r(t)$$

Basically you construct a v-t graph - which will be a curve.
The area under it is the displacement.
 
Last edited:

Related Threads on Distance travelled of an object falling from non-negligible height

Replies
2
Views
286
Replies
11
Views
946
Replies
3
Views
2K
Replies
6
Views
4K
Replies
21
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
14
Views
4K
Top