Calculating relativistic effects of motion in solar system

KenJackson
Messages
63
Reaction score
10
This question and answer are posed in Kim Stanley Robinson's novel "2312".

"Pauline, if someone had calculated the trajectory of an impactor to hit [an exact spot on the planet Mercury], but they forgot to include the relativistic precession of Mercury in their calculation and only used the classical calculus of orbital mechanics, how far would they miss by? Assume the impactor was launched from the asteroid belt a year earlier."

Pauline said, "The precession of Mercury is 5603.24 arc seconds per Julian century, but the portion of that caused by the curvature of space-time as described by general relativity is 42.98 arc seconds per century. Any trajectory a year in duration, plotted without that factored in, would therefore miss by 13.39 kilometers."

My question is, is this all made up? Or might it be accurate?
 
Physics news on Phys.org
Interesting. The numbers are in the novel the way I typed them.

Even though Pauline answered only the precession part of the question, I was fascinated by the possibility of the whole calculation. In recent years, NASA and other countries' space agencies have sent probes to Mars, asteroids and even a comet. But I think they all had thrusters to do course corrections along the way.

But if you throw a rock at Mercury from the asteroid belt, how would you calculate (even in theory) the direction and speed to make it hit a specific spot after a year's travel--with no mid-course corrections!?
 
KenJackson said:
In recent years, NASA and other countries' space agencies have sent probes to Mars, asteroids and even a comet. But I think they all had thrusters to do course corrections along the way.

They did, but the course corrections are very small; a probe can't carry enough fuel to make large course corrections, so it has to be launched very accurately to begin with. The course corrections are not always needed, but NASA allows for the possibility to be safe.

KenJackson said:
if you throw a rock at Mercury from the asteroid belt, how would you calculate (even in theory) the direction and speed to make it hit a specific spot after a year's travel--with no mid-course corrections!?

You would have to know the precise positions of the asteroid, the Sun, and the planets at launch, so you could compute the rock's trajectory to the required accuracy. It's tedious, but straightforward; you start out with the asteroid and Mercury moving in the field of the Sun as your first approximation, then just add in effects of other planets until you've taken into account every effect that's large enough to matter. Nowadays computers would do all the grunt work anyway.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...

Similar threads

Back
Top