Remember, Newton also said an apple also attract earth due to the gravitational acceleration of an apple but is so minuscule to be noticed and hence the falling of earth towards apple is ignored – so theoretically bowling ball and earth will strike first. - I can go into detail if the hint is not sufficient
Yes, I do apologise for not reading your OP carefully enough, but have you done the calculations?
Starting with:
Mass of earth = 5.9722 x 10
24 kg, mass of bowling ball = 7.26 kg and mass of apple = 0.10 kg.
Gravitational constant = 6.67408 x 10
-11m
3 kg
-1 s
-2, Radius of earth = 6,371,008 m.
Distance to fall = 10 m.
In the following, I have ignored any rotational component of the effective acceleration.
- For the bowling ball-earth combination:
I get, force = 71.2928655492049 N, acceleration of the ball = 9.81995393239737 m/s2 and of the earth = 1.1937 x 10-23 m/s2 and the time to fall 10 m = 1.42711927750553 s.
- For the apple-earth combination:
I get, force = 0.981995393239737 N, acceleration of the apple = 9.81995393239737 m/s2 and of the earth = 1.6443 x 10-25 m/s2 and the time to fall 10 m = 1.42711927750553 s.
I think we can safely neglect the acceleration of the earth whether they are dropped at the same time or separately.
When comparing a bowling ball and a feather, the traces of air left in the vacuum chamber would have a much greater effect.
PS: I hope I have made no mistakes.