When you drop a 0.38 kg apple, Earth exerts
a force on it that accelerates it at 9.8 m/s2
toward the earth’s surface. According to Newton’s third law, the apple must exert an equal but opposite force on Earth.
If the mass of the earth 5.98 × 1024 kg, what
is the magnitude of the earth’s acceleration
toward the apple?
Answer in units of m/s2
Answers
The magnitude of the earth’s acceleration toward the apple is 6.2 * 10⁻²⁵ m/s².
Explanation:
The mass of the apple, m = 0.34 kg
The acceleration of the apple towards the earth's surface, a = 9.8 m/s²
The mass of the Earth, M = 5.98 * 10²⁴ kg
Now, by applying Newton's Second Law of the motion,
The magnitude of the force on the apple due to Earth is,
F = ma
⇒ F = 0.34 * 9.8
⇒ F = 3.724 N
We are given that, according to Newton's third law of the motion, the apple must exert an equal but opposite force on Earth.
So, the magnitude of the force on the Earth due to the apple will be,
F = 3.724 N
Let the magnitude of the earth’s acceleration towards the apple be denoted as “A” m/s².
Thus,
The magnitude of the earth’s acceleration towards the apple will be given by:
F = MA
⇒ 3.724 = [5.8 * 10²⁴] * A
⇒ A = 3.724 / [5.8 * 10²⁴]
⇒ A = 0.6227 * 10⁻²⁴
⇒ A = 6.2 * 10⁻²⁵ m/s²
Hope this is helpful!!!!