Video Transcript
Consider a denser planet that has
the same volume as our planet but is four times the mass of Earth. A book is dropped from a height of
five meters to fall freely under the gravity of the planet. If the same book is dropped here on
Earth from the same height, find the ratio between the velocity just before it
reaches the ground here on Earth and the velocity just before it reaches the ground
on the other planet.
Recall that the acceleration due to
gravity 𝑎 at the surface of a uniform sphere of mass 𝑚 and radius 𝑟 is equal to
𝐺𝑚 over 𝑟 squared, where 𝐺 is the universal gravitational constant. This is the same formula as that of
a particle of mass 𝑚 located at the geometric center of the sphere.
We can model a planet at such a
uniform sphere. In this scenario, we have two
planets with the same radius 𝑟 but different masses, with the Earth of mass 𝑚 and
the other planet of mass four times this, four 𝑚. In the formula for acceleration due
to gravity, everything apart from the mass 𝑚 is therefore a constant. This therefore means that the
acceleration due to gravity 𝑎 is proportional to the mass 𝑚.
An increase in some factor of mass
will result in the same factor of increase of acceleration due to gravity. In this case, the increase in mass
is four times. Therefore, the acceleration due to
gravity on the other planet 𝑎 P is four times the acceleration due to gravity on
Earth, 𝑎 E.
Now, we need to use the kinematics
equations, also known as the SUVAT equations, to find the velocity of the book just
before it reaches the ground on both planets. The quantity we wish to find is the
final velocity 𝑣. We do not know the time it takes
the book to reach the ground in both scenarios. But we do have the acceleration due
to gravity, the initial velocity 𝑢, which is zero because the book starts from
rest, and the displacement 𝑠, which is five meters. The kinematics equation we need to
use then is the one which involves 𝑠, 𝑢, 𝑣, and 𝑎, which is 𝑣 squared equals 𝑢
squared plus two 𝑎𝑠.
The initial velocity 𝑢 is equal to
zero in both cases. Therefore, 𝑢 squared is also equal
to zero. We can therefore write 𝑣
explicitly as the square root of two 𝑎𝑠. We are asked to find the ratio
between the velocity just before the book reaches the ground on Earth, 𝑣 E, and the
velocity just before it reaches the ground on the other planet, 𝑣 P. 𝑣 E is equal to the square root of
two 𝑎 E 𝑠, and 𝑣 P is equal to the square root of two 𝑎 P 𝑠. We have a common factor on both
sides of the ratio of root two and root 𝑠. And since neither of these are
equal to zero, these will both cancel.
This demonstrates the interesting
property that the answer to this problem is actually independent of the distance
that the book falls from. So, the ratio 𝑣 E to 𝑣 P is equal
to the ratio root 𝑎 E to root 𝑎 P. 𝑎 P is equal to four times 𝑎
E. So, this ratio becomes root 𝑎 E to
root four 𝑎 E. We now have a common factor in both
sides of the ratio of 𝑎 E, which is also not equal to zero. So, these will both cancel. This leaves just one on the
left-hand side and the square root of four on the right-hand side.
Since both velocities are in the
same direction, vertically downwards, we can take this as the positive direction and
therefore ignore the negative square root, which gives us our final answer. The ratio between the velocity of
the book just before it reaches the ground here on Earth, 𝑣 E, and the velocity
just before it reaches the ground of the other planet, 𝑣 P, is equal to one to
two.
