Take the mass of the Earth to be 5.98 ×
1024 kg.
If the Earth’s gravitational force causes a
falling 74 kg student to accelerate downward
at 9.8 m/s2, determine the upward acceleration of the Earth during the student’s fall.
Answer in units of m/s2
Since the forces on student (due to Earth) and Earth (due to student)are equal and opposite,
F = m g = M a
a = (m/M)* = 1.24*10^-23 g
= ??
Two marbles, one twice as massive as the
other, are dropped from the same height.
When they strike the ground, how does the
kinetic energy of the more massive marble
compare to that of the other marble?
1. It has twice the KE.
2. It has 4 times the KE.
3. It has one-half the KE.
4. It has the same KE
To determine the upward acceleration of the Earth during the student's fall, we can use the principle of action-reaction.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the gravitational force that the Earth exerts on the student is equal in magnitude and opposite in direction to the gravitational force that the student exerts on the Earth while falling.
The gravitational force between two objects can be calculated using the formula:
F = (G * m₁ * m₂) / r²
Where:
F is the gravitational force
G is the gravitational constant (approximately 6.67 × 10^(-11) N m²/kg²)
m₁ and m₂ are the masses of the two objects
r is the distance between the centers of the two objects
In this scenario, we can consider the student and the Earth as the two objects. The mass of the student is 74 kg, and the mass of the Earth is 5.98 × 10^24 kg.
We know that the student is accelerating downward at 9.8 m/s² due to the Earth's gravitational force. We can use this information to calculate the magnitude of the gravitational force acting on the student:
F = m * a
Where:
F is the force
m is the mass of the student
a is the acceleration (-9.8 m/s², negative due to downward direction)
Substituting the values:
F = 74 kg * (-9.8 m/s²)
F = -725.2 N
Since the force between the student and the Earth is the same, but opposite in direction, the magnitude of the force between the Earth and the student is also 725.2 N.
Now, we can rearrange the gravitational force formula to solve for the acceleration (a) of the Earth:
a = (F * r²) / (G * m)
Where:
a is the acceleration of the Earth
F is the force between the student and the Earth
r is the distance between the centers of the student and the Earth (considered approximately the radius of the Earth, which is about 6.37 × 10^6 meters)
Substituting the values:
a = (725.2 N * (6.37 × 10^6 m)²) / (6.67 × 10^(-11) N m²/kg² * 5.98 × 10^24 kg)
a ≈ 0.01 m/s²
Therefore, the upward acceleration of the Earth during the student's fall is approximately 0.01 m/s².