Escape Velocity for Dummies

Calculating escape velocity

It's been almost six months since we last featured one of our most prolific DotDs, and a couple of months since we dipped our collective toe in the murky waters of HubPages.com and its special little niches¹. Mostly it's because searching the site tends to be a pain in the rear, not to mention that the place no longer pays jack so hubbies don't update their content and it tends to disappear... but we do make the occasional exception. Here without further ado, we give you 28-time winner Joan Whetzel and her take on "Escape Velocity and the Solar System."

Reading Joan's prose is always a trip, in no small part because she's even worse at proofreading herself than the average freelancer. That's how she ends up with sentences like,

"That gravitational force varies, depending on the of the [sic] planet or sun and the mass of the of the [sic] object to be launched into space as well as the square of the distance between the planet and the object to be launched."

Joanie's writing there (we think) about the force of gravity, so why she has to babble about "launched into space" is rather puzzling – almost as puzzling as "the mass of the of the object," we think. This is all from a rather clumsy attempt to put Newton's law of universal gravitation into words; an attempt that yielded,

"That force of gravity is equal to both masses (m1 = the mass of the planet, m2 = the mass of the person or object being sent into space) multiplied together, then divided by the distance between (r = the radius of the planet, squared) them [sic] squared. That number is then multiplied by the universal gravitational constant (G)²."

With the exception of the bull about "the radius of the planet," Joan's on the right track. What she apparently doesn't realize is that r is not "the radius of the planet." Instead, it's the distance between the two objects' centers. She's actually stating the force of gravity at the surface of a planet, although she doesn't seem to realize the difference.

Whetzel continues in that vein for several more paragraphs, creating mashups of various scientific terminology and concepts, as is her wont. It's what she penned in her section on the definition of escape velocity, however, that originally caught our research staffer's eye. According to Joan, the formula for escape velocity is,

V^e = [(2GM) / r]^²

There's just one little problem there: the formula is actually

V^e = [(2GM) / r]^½

Guess what: Joan's botched formula results in an answer that is incorrect by a whopping big factor. In fact, she ends up with the square of the real escape velocity multiplied by its square root! If you plug in the mass of the Earth (which Joan's table says is 5.98 x 10²⁴ kg) and the value of G, you end up with a number something like 420 km/sec, a far cry from the actual value of about 11.2 km/sec!
What a putz. What a classic Dumbass of the Day move.

¹ They include dengarden.com, AxleAddict.com, Owlcation.com...
² You'd think Joan would talk more about G, but she doesn't. It's approximately 6.674×10⁻¹¹ N·m²/kg² if you're interested – and you should be!

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