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If the atmosphere of an imaginary planet is really thin like 100 particles per cubic centimeter (density of the closest resemblance of vacuum made on Earth) and has mass say, 15 times than that of earth, does this planet fulfils the conditions to exhibit terminal velocity?

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From the equation given for terminal velocity

v=$$\sqrt{\frac{2mg}{\rho AC}}$$

the density of the planet in question will be very low , so any object dropped will take a very long time to acquire a terminal velocity. There are so little particles present in the atmosphere that the balancing of the weight by the drag and buoyancy will take a very long time .

If the height of the atmosphere is enough to get the velocity of the object dropped to the terminal velocity then it will or it wont and just keep accelerating till it hits the ground or reaches the center of the planet in case there is no solid ground in the planet.

Hope my answer helps

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  • $\begingroup$ what does $C_d$ in that formula means $\endgroup$ – Dheeraj Kumar Dec 13 '14 at 11:15
  • $\begingroup$ Its the drag coefficient. can read more about it here en.wikipedia.org/wiki/Drag_coefficient The drag coefficient is a dimensionless quantity that is used to quantify the drag or resistance of an object in a fluid environment $\endgroup$ – Gowtham Dec 13 '14 at 11:50
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The air density of your planet is on the order of $10^{-18}$ of air at the surface of the earth. Gravity Force would be 10 times larger, even accounting for a slightly larger radius. The largest possible parachute that could fit around such a planet would be in the order of $10^{12}$ times bigger than a normal parachute. But it would need to have a mass in milligrams (at most), which is not enough mass for even a single atom thick string of carbon atoms around the circumference of our parachute.

Basically, any material that is itself not a gas would by many orders of magnitude too heavy to be even slightly affected by the air in your world, let alone reach terminal velocity.

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