The Motion Of Objects Thrown Away From Earth

  • Thread starter Thread starter Correia
  • Start date Start date
  • Tags Tags
    Earth Motion
AI Thread Summary
The discussion focuses on deriving the distance function for an object thrown away from Earth, using the gravitational acceleration formula g = c/r^2. A proposed approach involves solving the differential equation r'' = c/r^2, although the original poster lacks experience with differential equations. Participants suggest using a spherical polar coordinate system and constructing a free body diagram to apply Newton's second law. Clarifications are sought regarding whether the object is thrown directly upwards or in a different trajectory, as this affects the calculations. The conversation emphasizes the need to consider the Earth's rotation and other factors in the analysis.
Correia
Messages
1
Reaction score
0
How can the function for distance over time be found for an object which is thrown away from Earth? Acceleration as a function of distance is

g = GM/r^2,

where GM is constant, so let c = GM, then

g = c/r^2.

Then I wondered, maybe the formula for distance can be found by solving the differential equation

r'' = c/r^2?

I have never studied differential equations, so I have no idea about it, even to whether I can really say r'' = c/r^2.

If someone can elucidate my mind, I shall be grateful.
 
Physics news on Phys.org
Welcome to PF;

Have a look at:
http://home.comcast.net/~szemengtan/ClassicalMechanics/SingleParticle.pdf
The general approach is given in section 1.7: "Central Force Problem"

The basic technique should be familiar, you choose a coordinate system that makes sense (spherical polar) and construct a free body diagram for your mass... put ##\vec{F}=m\vec{a}## and solve the differential equation for the appropriate initial condition.

Were you thinking of the Earth specifically or a non-rotating spherical mass M radius R?
Did you intend to throw the mass m directly upwards? (i.e. radially outwards) or solve the ballistics problem for the situation where the projectile does not remain close to the surface?
(The rotation of the Earth, etc, affects the result.)

Note: ##\vec{F}=-mg(r)\vec{r}/r: r>R##
 
Last edited by a moderator:
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top