Question

Two rocks (call them S and T) are released at the same time from the same height and start from rest. Rock S has 20 times the mass of rock T. Which rock will fall faster if the only forces involved are each rock’s mutual gravitational attraction with Earth?

Answer 1

Answer:At same time

Step-by-step explanation:

Given

Rock s has 20 times the mass of rock T

They are released from the same height

suppose height from which they are released is h with initial velocity =0

`h=ut+(gt^2)/(2)`

`h=(gt^2)/(2)`

`t=\sqrt{(2h)/(g)}`

as we can see from above expression that time is independent of mass therefore they both reach at same time

Answer 2

Both will fall at the same speed

Step-by-step explanation:

Both rocks S and T will reach the ground at the same time this means the velocities they reach at the end of the fall will be equal assuming that their initial velocities are same and are released at the same time. This happens due to the fact that the Earth's gravity causes an acceleration that is equal irrespective of the mass of the object.

m = Mass of object

M = Mass of the Earth = 5.972 × 10²⁴ kg

r = Radius of Earth = 6371000 km

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

`F=ma`

`F=G(Mm)/(r^2)`

`\\\Rightarrow ma=G(Mm)/(r^2)\\\Rightarrow a=G(M)/(r^2)\\\Rightarrow a=6.67* 10^(-11)(5.972* 10^(24))/((6371* 10^3+6400)^2)\\\Rightarrow a=9.79395\ m/s^2`

The acceleration due to gravity on any body will be 9.79395 m/s²