Space Probes & Slingshot Effect: Energy Source & Violation of Laws?

  • Thread starter Thread starter simon009988
  • Start date Start date
  • Tags Tags
    Slingshot
AI Thread Summary
The slingshot effect allows space probes to gain speed by utilizing the gravitational pull of a planet, with the extra energy coming from the planet itself. This energy transfer is negligible due to the vast difference in mass between the planet and the probe, similar to how a butterfly's flight has an imperceptible effect on Earth's rotation. As the probe enters the planet's gravitational well, it is accelerated by the planet's orbital motion, which contributes to its increased velocity upon exit. While the probe gains speed, it can also lose speed under specific conditions during the flyby. Overall, the slingshot effect adheres to the laws of thermodynamics, as the energy exchange is balanced and mathematically consistent.
simon009988
Messages
51
Reaction score
0
when space probes pass planets and gains speed via the slingshot effect; where does the extra energy come from? will it not violate the laws of thermodynamics? I know it doesn't but why?
 
Physics news on Phys.org
The extra energy comes from the planet it's passing. It will be essentially umeasurable since the planet is so much bigger than the probe.
 
If a butterfly is flying against the Earths rotation it will have the influence of slowing the Earths rotation by the Butterfly's drag coefficient while flying on the micro scale, the spin loss is so small that it would be difficult to detect but not mathmatically, but the reverse can also happen in which if the Butterfly is traveling with the Earth's rotation.
 
As the probe falls into the planets gravitational well it is pulled along with the planet in its orbital motion. It is this extra velocity that the probe takes with it as it leaves the planets gravitational well. All velocity gained due to the gravatational well is lost on exit, the orbital velocity remains, this is the slingshot effect.
 
BTW - it is entirely possible for a spacecraft to lose speed as it passes by a planet. You need to set up the right conditions for it to acquire energy.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top