• Support PF! Buy your school textbooks, materials and every day products Here!

Gravitational Law - Very Difficult

  • Thread starter mitch_nufc
  • Start date
  • #1
An object is fired vertically upwards from the surface of a planetary body; it moves under the action of Newton’s Gravitational Law, without
resistance, so the equation is z'' = -gR^2 / (z + R)^2 . Find the relation between v = z' and z and use this model, and the relation that you have
found, to obtain a numerical estimate for the escape speed on the surface of the Earth.

Its the wording of the question I don't get? I assume I integrate z'' twice to find the velocity and distance respectively, but i already have a z on the RHS so i cant integrate, but the escape speed to escape the surface of the earth? is this like when g (9.81) begins to decrease as it moves from the planetry body? I havn't got a clue at all. Any help appreciated
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
The escape velocity is the velocity at which the kinetic energy is equal to the gravitational potential energy of the planet.

Using the chain rule you can write [tex]\frac{d^2 z}{dt^2}=\frac{d v}{dt}=\frac{dv}{dz}\frac{dz}{dt}=v\frac{dv}{dz}[/tex] After rearranging your equation you can now integrate.

How did you get the equation for z''?
 
  • #3
it was given in the question, the equation for z'' that is
 
  • #4
Cyosis
Homework Helper
1,495
0
Odd that the mass of the planet isn't a part of the equation.
 
  • #5
Does it help to know I'm doing a Maths degree and not a Physics one? We were told none of our questions would require knownledge of classical mechanics/physics, but just differential equations etc
 
  • #6
Cyosis
Homework Helper
1,495
0
Regardless of what they said this is somewhat of a physics problem, but it matters not. Just use the equation you were given.
 
  • #7
472
0
Hi there,

Your first equation: [tex]\frac{d^2z}{dt^2} = ...[/tex] is not Newton's gravitational law. It is the result of the Newton's gravitational law, considering a planetary body, like the Earth, the Sun, or the Moon. Newton's real gravitational law is define as the force that attracts two body of masses, and is express with : [tex]F_g = G\frac{m_1 \cdot m_2}{r^2}[/tex] where [tex]G[/tex] is Universal gravitational constant, [tex]m_i[/tex] are the mass of each body, and [tex]r[/tex] is the distance between the two body's center of mass.

Hope this helps you understand it more.
 
  • #8
yeah i remember all this from A-level physics, but i dont know any masses or distances so i assume theyre kept as constants in the ODE
 
  • #9
107
1
There is no mention to the mass of the planet because it's inside g, the formula is correct (physically speaking). If this was merely a maths problem, you'd have to solve the ODE with the initial conditions z(0)=0, z'(0)=v and find out for which v z(infty)->infty. But this is too much work for a physicist... We know that there is a conserved quantity, energy:

[tex]E=z'^2 - {gR^2\over z+R}[/tex]

and that for E=0, the particle will escape, so... we set z=0 in that equation and solve for z'
 

Related Threads for: Gravitational Law - Very Difficult

Replies
0
Views
1K
  • Last Post
Replies
7
Views
1K
Replies
1
Views
456
Replies
0
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
4
Views
3K
Top