Skip to main content
added 1456 characters in body
Source Link
Thomas Pornin
  • 1.1k
  • 7
  • 13

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse.

A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is the "lowest orbit" because the satellite grazes the ground -- we ignore the atmosphere here).

The oscillation in the hole is really another orbit -- it is a degenerate ellipse which has been flattened to a line. Yet its semi-major axis is still the Earth radius.

Same semi-major axis, hence same period.

Edit: as was pointed out, that expanation is bogus in two ways:

  • The degenerate case for a "flattened" ellipse would be a half-diameter. If all the Earth's weight was concentrated at its center, the orbit, starting from "ground" level (6300 km or so from the center) with (almost) no lateral velocity would be an accelerated fall toward the center; when close to the center, the object would miss it "by mere inches" and quickly run around it, before speeding up back to the initial position at ground level. Furthermore, that "flattened ellipse" would have a semi-major axis of length about 3150 km (half the radius), for a period which would be eight times smaller than the low orbit.

  • The Earth weight is not concentrated at its center. In fact you get an "oscillator" trajectory, that allows you to emerge in New Zealand if you started from England, precisely because the "Earth mass at a single point" model is not the one used in this thought experiment.

While it is understandable that the low orbit and the ocsillator end up with periods of the same magnitude (they both are kinds of "free fall" against an Earth with the same weight, and starting at ground level), that hand-waving remark would have been equally applicable with an oscillator period being twice or half that of the low orbit. They seem to end up quite close to each other and I now have no idea whether this is mere coincidence or for some fundamental reason.

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse.

A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is the "lowest orbit" because the satellite grazes the ground -- we ignore the atmosphere here).

The oscillation in the hole is really another orbit -- it is a degenerate ellipse which has been flattened to a line. Yet its semi-major axis is still the Earth radius.

Same semi-major axis, hence same period.

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse.

A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is the "lowest orbit" because the satellite grazes the ground -- we ignore the atmosphere here).

The oscillation in the hole is really another orbit -- it is a degenerate ellipse which has been flattened to a line. Yet its semi-major axis is still the Earth radius.

Same semi-major axis, hence same period.

Edit: as was pointed out, that expanation is bogus in two ways:

  • The degenerate case for a "flattened" ellipse would be a half-diameter. If all the Earth's weight was concentrated at its center, the orbit, starting from "ground" level (6300 km or so from the center) with (almost) no lateral velocity would be an accelerated fall toward the center; when close to the center, the object would miss it "by mere inches" and quickly run around it, before speeding up back to the initial position at ground level. Furthermore, that "flattened ellipse" would have a semi-major axis of length about 3150 km (half the radius), for a period which would be eight times smaller than the low orbit.

  • The Earth weight is not concentrated at its center. In fact you get an "oscillator" trajectory, that allows you to emerge in New Zealand if you started from England, precisely because the "Earth mass at a single point" model is not the one used in this thought experiment.

While it is understandable that the low orbit and the ocsillator end up with periods of the same magnitude (they both are kinds of "free fall" against an Earth with the same weight, and starting at ground level), that hand-waving remark would have been equally applicable with an oscillator period being twice or half that of the low orbit. They seem to end up quite close to each other and I now have no idea whether this is mere coincidence or for some fundamental reason.

Source Link
Thomas Pornin
  • 1.1k
  • 7
  • 13

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse.

A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is the "lowest orbit" because the satellite grazes the ground -- we ignore the atmosphere here).

The oscillation in the hole is really another orbit -- it is a degenerate ellipse which has been flattened to a line. Yet its semi-major axis is still the Earth radius.

Same semi-major axis, hence same period.