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A student of mass 45 kg jumps off a high diving board. What is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of 9.8 m/s\(^2\)? Use 6.0 \(\times\) 10\(^{24}\) kg for the mass of the earth, and assume that the net force on the earth is the force of gravity she exerts on it.

Short Answer

Expert verified
The Earth's acceleration toward the student is approximately \( 7.35 \times 10^{-23} \text{ m/s}^2 \).

Step by step solution

01

Understand Newton's Third Law

According to Newton's Third Law of Motion, when the student jumps off the diving board, she exerts a gravitational force on the Earth, and the Earth exerts an equal and opposite force on her. Therefore, the force exerted by the student is equal to the gravitational force acting on her.
02

Calculate the Gravitational Force Exerted by the Student

The force of gravity acting on the student can be calculated using the formula: \[ F = m \cdot g \]where \( F \) is the force, \( m \) is the mass of the student (45 kg), and \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)).Substituting the given values:\[ F = 45 \text{ kg} \times 9.8 \text{ m/s}^2 = 441 \text{ N} \]
03

Apply Newton's Second Law to the Earth

Newton's second law states that \( F = M \cdot a \), where \( F \) is the force, \( M \) is the mass of an object, and \( a \) is its acceleration. We need to find the acceleration of the Earth (\( a_e \)) due to the gravitational force exerted by the student.
04

Calculate the Earth's Acceleration

Using the gravitational force calculated earlier, apply it to the equation \[ F = M \cdot a_e \]where \( F = 441 \text{ N} \) is the force, \( M = 6.0 \times 10^{24} \text{ kg} \) is the mass of the Earth. Solve for \( a_e \):\[ 441 \text{ N} = (6.0 \times 10^{24} \text{ kg}) \times a_e \]\[ a_e = \frac{441}{6.0 \times 10^{24}} \text{ m/s}^2 \]\[ a_e \approx 7.35 \times 10^{-23} \text{ m/s}^2 \]
05

Conclusion

The acceleration of the Earth toward the student is approximately \( 7.35 \times 10^{-23} \text{ m/s}^2 \). This value is extremely small due to Earth's large mass.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another. In simpler terms, gravity is the force that causes objects to pull toward each other. Whenever you jump, you come back down because Earth's gravitational force is pulling you. This same force keeps the planets in orbit around the sun and the moon orbiting Earth.

The strength of gravitational force depends on two key factors:
  • Mass of the objects involved: Larger masses exert a stronger gravitational pull.
  • Distance between the centers of the two objects: Greater distances result in weaker gravitational forces.
When you jump or drop something, gravity acts as an accelerating force toward the center of the Earth. This acceleration due to gravity is usually represented by the symbol \( g \) and has a standard value of \( 9.8 \text{ m/s}^2 \) on Earth. This is why objects fall towards the ground if not supported.
Acceleration
Acceleration refers to the rate of change of velocity of an object with respect to time. It signifies how fast an object is speeding up or slowing down and is a vector quantity, meaning it has both magnitude and direction.

Newton's Second Law of Motion connects acceleration (\( a \)) with force (\( F \)) and mass (\( m \)) using the formula: \[ F = m \cdot a \]This formula implies that:
  • The acceleration of an object is directly proportional to the force acting upon it.
  • Acceleration is inversely proportional to the mass of the object.
So, heavier objects need more force to achieve the same acceleration as lighter ones. In our example, the student jumping off the diving board experiences gravity causing her to accelerate downwards at \( 9.8 \text{ m/s}^2 \), typical of any free-falling body near Earth.

The relationship between acceleration and the forces acting on objects is crucial in understanding motion, and it is especially pivotal when calculating the Earth's slight acceleration in response to forces applied to it.
Mass of the Earth
The mass of the Earth is an incredibly large value, approximately \( 6.0 \times 10^{24} \text{ kg} \). This massive quantity is pivotal in calculations involving gravitational force and planetary motion. One way to visualize such a large number is by considering the weight of the Earth if it were placed on a giant balance scale. Its immense size and mass mean that Earth exerts a strong gravitational pull, which is why it influences the motion of nearby bodies like the moon and satellites.

An important point to note is that the Earth's enormous mass results in very minimal acceleration when subjected to normal forces. This concept was demonstrated in the exercise example, where the planet's acceleration towards a student jumping off a diving board was almost negligible: \( 7.35 \times 10^{-23} \text{ m/s}^2 \).
  • Such a tiny acceleration is due to the Earth’s vast mass compared to the small external force applied by an individual student.
  • Therefore, while any force applied has a theoretical impact, the Earth's movement is virtually imperceptible due to its hugeness.
Understanding Earth's mass helps to grasp the scope of gravitational influence and its effect on both celestial and terrestrial mechanics.

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Most popular questions from this chapter

An object of mass \(m\) is at rest in equilibrium at the origin. At \(t =\) 0 a new force \(\vec{F}(t)\) is applied that has components $$F_x(t) = k_1 + k_2y$$ $$F_y(t) = k_3t$$ where \(k_1\), \(k_2\), and \(k_3\) are constants. Calculate the position \(\vec{r}(t)\) and velocity \(\vec{v}(t)\) vectors as functions of time.

A 75.0-kg man steps off a platform 3.10 m above the ground. He keeps his legs straight as he falls, but his knees begin to bend at the moment his feet touch the ground; treated as a particle, he moves an additional 0.60 m before coming to rest. (a) What is his speed at the instant his feet touch the ground? (b) If we treat the man as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant? (c) Draw his freebody diagram. In terms of the forces on the diagram, what is the net force on him? Use Newton's laws and the results of part (b) to calculate the average force his feet exert on the ground while he slows down. Express this force both in newtons and as a multiple of his weight.

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To study damage to aircraft that collide with large birds, you design a test gun that will accelerate chicken-sized objects so that their displacement along the gun barrel is given by \(x =\) (9.0 \(\times\) 10\(^3\) m/s\(^2)t^2\) - (8.0 \(\times\) 10\(^4\) m/s\(^3)t^3\). The object leaves the end of the barrel at \(t =\) 0.025 s. (a) How long must the gun barrel be? (b) What will be the speed of the objects as they leave the end of the barrel? (c) What net force must be exerted on a 1.50-kg object at (i) \(t =\) 0 and (ii) \(t =\) 0.025 s?

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