Take the mass of the Earth to be 5.98 × 1024 kg.
If the Earth’s gravitational force causes a falling 44 kg student to accelerate downward at 9.8 m/s2, determine the upward accelera- tion of the Earth during the student’s fall.
Answer in units of m/s2.
LOL
F = G Me m/r^2
F/m = G Me/r^2 = 9.8
F/Me = G m/r^2 = 9.8 (m/Me)
so
9.8 (44/5.98*10^24)
Well, you know what they say: what goes up must come down, even for Earth! So, let's calculate the upward acceleration of Earth during the student's fall.
To find the upward acceleration, we can use Newton's third law, which states that for every action, there is an equal and opposite reaction. In this case, the student is exerting a downward force on the Earth, and the Earth exerts an equal and opposite upward force on the student.
Given that the student's mass is 44 kg and the downward acceleration is 9.8 m/s², we can calculate the force exerted by the student on Earth using Newton's second law: force = mass × acceleration.
Force = (44 kg) × (9.8 m/s²) = 431.2 N
Now, since the forces are equal and opposite, the upward force exerted by the Earth on the student is also 431.2 N.
Using Newton's second law again, we can find the upward acceleration of the Earth:
Force = mass × acceleration.
431.2 N = (5.98 × 10²⁴ kg) × acceleration
Shuffle things around a bit and solve for acceleration, and we get:
acceleration = 431.2 N / (5.98 × 10²⁴ kg) ≈ 7.203 × 10⁻²⁴ m/s²
So, the upward acceleration of the Earth during the student's fall is approximately 7.203 × 10⁻²⁴ m/s².
Hope that answer didn't bring you down too much!
To determine the upward acceleration of the Earth during the student's fall, we can use Newton's third law of motion. According to Newton's third law, the force of the Earth pulling the student downward is equal in magnitude and opposite in direction to the force of the student pulling the Earth upward.
We know the mass of the Earth (M = 5.98 × 10^24 kg) and the mass of the student (m = 44 kg). We also know the acceleration of the student (a = 9.8 m/s^2).
Let's calculate the upward acceleration of the Earth using the following formula:
F = ma
Where F represents the force and a represents the acceleration.
The force of the Earth pulling the student downward is given by:
F = mg
Where g represents the gravitational acceleration on the surface of the Earth (approximately 9.8 m/s^2).
Substituting the values into the formula:
F = (44 kg) * (9.8 m/s^2)
F = 431.2 N
According to Newton's third law, the force of the student pulling the Earth upward is also 431.2 N. Therefore, the upward acceleration of the Earth can be calculated using the formula mentioned earlier:
a = F / M
Substituting the values into the formula:
a = (431.2 N) / (5.98 × 10^24 kg)
a ≈ 7.2 × 10^-23 m/s^2
So, the upward acceleration of the Earth during the student's fall is approximately 7.2 × 10^-23 m/s^2.
To determine the upward acceleration of the Earth during the student's fall, we need to use the concept of Newton's third law of motion. According to Newton's third law, every action has an equal and opposite reaction.
In this case, the action is the gravitational force of the Earth pulling the student downward, and the reaction is the equal magnitude force with the opposite direction acting on the Earth. The force acting on the Earth is given by:
Force = mass * acceleration
Using the values given in the question:
Mass of the Earth (M) = 5.98 × 10^24 kg
Mass of the student (m) = 44 kg
Acceleration of the student (a) = 9.8 m/s^2
We can calculate the force acting on the student using Newton's second law:
Force = m * a
Force = 44 kg * 9.8 m/s^2
Force = 431.2 N
Since the gravitational force acting on the student is equal in magnitude and opposite in direction to the reaction force acting on the Earth, the force acting on the Earth is also 431.2 N.
Now, to calculate the upward acceleration of the Earth (a_up), we can rearrange the formula:
Force = M * a_up
Rearranging the formula:
a_up = Force / M
Plugging in the values:
a_up = 431.2 N / 5.98 × 10^24 kg
Now, let's calculate the value:
a_up = 7.2 × 10^(-23) m/s^2
Therefore, the upward acceleration of the Earth during the student's fall is 7.2 × 10^(-23) m/s^2.