A student with mass 55 kg jumps off a high diving board.
Using 6.0×10^24 kg for the mass of the earth, what is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of 9.8 m/s2 ? Assume that the net force on the earth is the force of gravity she exerts on it.
F = 6.0x10^24 kg x 9.8 m/s^2
F = 5.88x10^25 N
5.88x10^25 N = 55 kg * a
a = 1.07 x^24 m/s^2
The acceleration of the earth toward the student is 1.07 x 10^24 m/s^2.
To find the acceleration of the Earth toward the student, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
Given that the mass of the Earth is 6.0×10^24 kg and the student's acceleration toward the Earth is 9.8 m/s^2, we can calculate that the force exerted by the student on the Earth is:
F = (6.0×10^24 kg) * (9.8 m/s^2)
= 5.88×10^25 kg·m/s^2
= 5.88×10^25 N
Now, since the net force on the Earth is the force of gravity that the student exerts on it, we can equate this force to the mass of the student multiplied by the acceleration of the Earth:
5.88×10^25 N = (55 kg) * a
Rearranging the equation, we can solve for the acceleration of the Earth:
a = (5.88×10^25 N) / (55 kg)
= 1.07×10^24 m/s^2
Therefore, the acceleration of the Earth toward the student is approximately 1.07×10^24 m/s^2.
To find the acceleration of the earth towards the student, we can use Newton's law of universal gravitation, which states that the force of gravitational attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers of mass.
In this case, we need to use the equation:
F = G * ((m1 * m2) / r^2)
Where:
- F is the force of gravitational attraction between the student and the Earth
- G is the gravitational constant, approximately 6.67430 × 10^-11 N(m/kg)^2
- m1 is the mass of the student
- m2 is the mass of the Earth
- r is the distance between the student and the Earth's center of mass (which we can consider as the radius of the Earth)
The mass of the Earth is given as 6.0×10^24 kg, the acceleration of the student towards the Earth is given as 9.8 m/s^2, and the mass of the student is given as 55 kg.
First, let's calculate the force of gravitational attraction between the student and the Earth:
F = G * ((m1 * m2) / r^2)
F = (6.67430 × 10^-11 N(m/kg)^2) * ((55 kg) * (6.0×10^24 kg)) / (r^2)
Since we want to find the acceleration of the Earth towards the student, we need to rearrange the equation:
F = m2 * a
a = F / m2
Substituting the values:
a = (6.67430 × 10^-11 N(m/kg)^2 * ((55 kg) * (6.0×10^24 kg)) / (r^2)) / (6.0×10^24 kg)
Simplifying the equation, we find:
a = (6.67430 × 10^-11 N(m/kg)^2 * (55 kg)) / (r^2)
Now, let's calculate the acceleration using the calculated gravitational force:
F = 5.88x10^25 N (from the given calculation in the question)
r = radius of Earth = 6.37 × 10^6 meters
a = (6.67430 × 10^-11 N(m/kg)^2 * (55 kg)) / (6.37 × 10^6 meters)^2
Calculating the value, we find:
a ≈ 1.07 × 10^-24 m/s^2
Therefore, the acceleration of the Earth towards the student is approximately 1.07 × 10^-24 m/s^2.