According to the nursery rhyme, the cow jumped over the moon. But coming back, man, that would not be good.
Let’s assume the cow’s trajectory takes it in a free return trip around the Moon, a la Apollo 13. (Apollo 8 slowed and orbited the Moon ten times.) Let’s also assume our cow is wearing a spacesuit with days of oxygen so no problem there. So I assume that at some point after departing the Moon, the cow will pass from the Moon’s gravitational influence and start accelerating towards the Earth.
I cannot figure out how fast the cow will be going when it slams into the Earth’s atmosphere. I’m not smart enough. The 9.8msq/second thing is the rate of acceleration at sea level, which obviously ain’t true a hundred thousand miles out. Is anyone here smart enough to come up with a rough estimate of how fast Bessie will be going before becoming beef jerky in the upper atmosphere?
I don’t see how the cow escapes the Moon’s gravitational pull to start falling towards Earth: If she jumps to the Moon, she stays around the Moon.
That is, assuming all three bodies, Earth, Moon, and Cow are roughly spherical.
She escaped Earth’s gravity by jumping, that is exerting a force against the surface of the Earth. She cannot do that in space. And she jumped over the Moon, not to the Moon and from there back to Earth again.
That does not count as jumping in my book. That is rocket science!
Since no cow could actually generate enough speed by jumping to escape Earth’s gravity and would burn up before leaving the atmosphere, magic is involved. Magic solves all problems.
That’s my answer, too, though probably just a touch less. Since the cow has no means of propulsion in space, in order to jump over the moon she’d have to propel herself off the earth with enough speed to reach the L1 earth-moon Lagrange point, at which point the moon’s gravity takes over. L1 is around 85% of the distance to the moon, so the required speed of the jump would be fairly close to escape velocity, but not much more or Bessie would go flying past the moon and never come back. She’ll need to be traveling slow enough for the moon’s gravity to flip her around but fast enough to reach L1 again and start accelerating toward earth. From that distance, the unfortunate cow would hit the atmosphere at something fairly close to escape velocity.
According to my (admittedly very vage) mental calculations this orbit would pass below the Moon’s surface, that is, the cow would crash on the Moon, not turn around the Moon. If she avoids crashing on the Moon the trajectory is no longer elliptical but hyperbolic, then she misses the Earth on her way back. As a matter of fact she misses the Earth anyway, because during the cow’s coasting in space the Earth has moved along its orbit.
Because she would have to use a trajectory that is so close to the Moon’s centre of mass that it would run below the surface of the Moon at the turning point, so the cow would crash against the Moon. And then she would no longer be spherical!
Or if you prefer a pseudo-scientific approach: perhaps the cow has managed to figure out an extra, previously unsuspected, term to Einstein’s tensor gravity equation?
Better than ending up as steaks, I guess. By the way how does the cow breath in space?
I could be wrong but I believe that the Apollo missions flew to the moon on a free-return (ballistic) trajectory as a safety precaution, most notably Apollo 13 which made use of it (and so did the Soviet Luna 3 mission). For sure you can’t get into orbit around the earth or any other body through purely ballistic means from the surface without crashing back into it, but I don’t believe this applies to a ballistic trajectory where a spacecraft or cow departs a large body like earth and is deflected back by the gravity of a smaller body like the moon.
Although the earlier Apollo missions used a free-return trajectory (for cislunar mission often referred to as a “figure eight” trajectory), Apollo 12 and on did not in order to reduce transit time, instead using a slightly faster hybrid trajectory requiring small injection and return burns by the Service Module SPS engine. This is why Apollo 13 had to perform correction buns with the Lunar Module descent engine (unlike the Ron Howard film, multiple times) to align the return trajectory. 11.1 km/sec is about the minimum speed for a circumlunar lunar free-return although if the cow is very patient and well trained in celestial mechanics there are low energy transfers that might come in slightly slower.
That is incorrect. She would need to slow down at the moon in order to stay in orbit around the moon. Otherwise, she’s headed back to the Earth, or shooting off away from everything depending on how she approached the moon. To accurately answer the OP, we need to know a lot more information. An elliptical orbit going around the moon and back to Earth could result in a wide variety of velocities. There is a lower and upper bound, sure, but there’s a huge gap in the middle. The result is the same, no matter what, though. The cow doesn’t survive.
If the characteristic energy C3 of the cow is negative, i.e. the cow is not on an Earth escape trajectory, the return speed of the cow can’t be more than 11.2 km/s, and a trajectory returning from beyond the Moon can’t be much less. Of course, the cow could go into a heliocentric orbit that just happens to pass by the Moon and which eventually comes back to intercept the Earth, and could have a speed up to just under 42.1 km/s but that would be an extraordinarily patient cow.