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An experiment is performed in deep space with two uniform spheres, one with mass 50.0 kg and the other with mass 100.0 kg. They have equal radii, \(r =\) 0.20 m. The spheres are released from rest with their centers 40.0 m apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are 20.0 m apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the 50.0-kg sphere do the surfaces of the two spheres collide?

Short Answer

Expert verified
Linear momentum is conserved due to no external forces. Calculate using conservation laws. Use relative distances and velocities for specifics.

Step by step solution

01

Understanding Conservation of Linear Momentum

In a system with no external forces, the linear momentum of the system remains conserved. In this problem, since the only force acting is the mutual gravitational attraction between the two spheres and there are no external forces, the linear momentum of the system is conserved.
02

Initial Parameters and Gravitational Force

Consider the gravitational force between the two spheres as they move closer. Since they start from rest, their initial velocities are 0, and as they move closer, the force of attraction increases causing them to accelerate towards each other.
03

Conservation of Momentum Equation

Use the conservation of momentum: since the initial momentum of the system is zero (because both spheres are initially at rest), the momentum at any other time must also be zero.\[m_1 v_1 + m_2 v_2 = 0\]where \(m_1 = 50 \text{ kg}, m_2 = 100 \text{ kg}\), and \(v_1\) and \(v_2\) are the velocities.
04

Calculate Velocities When Centers are 20.0 m Apart

Using the conservation of energy, convert the initial potential energy at 40 m separation to kinetic energy at 20 m separation:Initial potential energy, \[U_i = -\frac{G m_1 m_2}{r_i}\]where \(r_i = 40\) m. Final potential energy,\[U_f = -\frac{G m_1 m_2}{r_f}\]where \(r_f = 20\) m. Kinetic energy at 20 m,\[K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\]Set initial potential energy equal to the sum of final potential energy and kinetic energy, solve for \(v_1\) and \(v_2\), respecting conservation of momentum to find individual velocities.
05

Calculate Relative Velocity

Relative velocity between the two spheres is given by\[v_{\text{relative}} = v_1 + v_2\]Substitute the velocities obtained from Step 4 to find the relative velocity.
06

Determine Collision Distance from Initial Position

Calculate where the centers are when surfaces collide:The separation between the centers when the surfaces touch is twice the radius,\[d = 2r = 0.4 \text{ m}\]Using the momentum (or center of mass) approach, determine the distance each moved. Since the center of mass (CM) remains at the same place, solve the equation for the distance moved by the 50 kg sphere from its initial position:\[s_1 = \frac{m_2}{m_1 + m_2} \times \text{distance moved by CM}\]

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Key Concepts

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

Conservation of Momentum
In physics, the concept of conservation of momentum states that if no external forces are acting on a system, the total momentum of that system remains constant over time. Momentum, which is the product of an object's mass and its velocity, is a vector quantity, meaning it has both magnitude and direction. In the given exercise, the only force acting is the gravitational attraction between the two spheres, and no outside forces are influencing their movement.

Thus, when the two spheres begin to move toward each other, their individual momentums change as they accelerate, but their combined momentum remains zero, as they started from rest. The equation for momentum conservation \(m_1 v_1 + m_2 v_2 = 0\) reflects this. Here, \(m_1\) and \(m_2\) are the masses of the spheres, and \(v_1\) and \(v_2\) are their velocities. Because the initial momentum was zero, the spheres always maintain opposite and equal momentum.
Conservation of Energy
Conservation of energy is a fundamental principle stating that energy in an isolated system remains constant. In this scenario, it means the total mechanical energy (sum of potential and kinetic energy) remains unchanged as the spheres move closer. Initially, the spheres possess only gravitational potential energy due to their separation. As they attract each other, this potential energy converts into kinetic energy.

To quantify this, we initially calculate the potential energy with the formula \[U_i = -\frac{G \cdot m_1 \cdot m_2}{r_i}\] where \(G\) is the gravitational constant, and \(r_i\) is the initial distance. When they are 20 m apart, their final potential energy changes, and the gained kinetic energy is given by \[K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\]. By equating the initial and final total energies, we can determine velocities \(v_1\) and \(v_2\). This approach ensures the energy transformation is balanced.
Gravitational Force Calculation
Gravitational force calculations are vital to understand how the spheres move towards each other. The law of universal gravitation is employed here, which states that every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The force \(F\) between the two spheres when they are at a distance \(r\) is calculated using \[F = \frac{G m_1 m_2}{r^2}\], where \(G\) is the gravitational constant. In the given problem, as the spheres start at 40 m apart and converge to 20 m apart, the gravitational force between them increases significantly. This increase in force translates into more acceleration, thus increasing the spheres' velocities as they draw near.
Relative Velocity in Physics
Relative velocity refers to the velocity of an object as observed from the reference frame of another object. In physics, understanding relative velocity is important when analyzing motion from different perspectives.

In the exercise, once we know the individual velocities of the spheres \(v_1\) and \(v_2\), the relative velocity is simply the vector sum of these velocities: \[v_{\text{relative}} = v_1 + v_2\]. This gives us the speed at which one sphere is approaching the other, taking into account their respective directions and speeds. The concept helps in visualizing the approach speed of the spheres, helping in understanding how quickly they close the distance between them.

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Most popular questions from this chapter

A uniform sphere with mass 50.0 kg is held with its center at the origin, and a second uniform sphere with mass 80.0 kg is held with its center at the point \(x =\) 0, \(y =\) 3.00 m. (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 kg placed at the point \(x =\) 4.00 m, \(y =\) 0? (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.0 \(\times\) 10\(^4\) m/s when at a distance of 2.5 \(\times\) 10\(^{11}\) m from the center of the sun, what is its speed when at a distance of 5.0 \(\times\) 10\(^{10}\) m?

The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s\(^2\). Neptune has mass 1.02 \(\times\) 10\(^{26}\) kg and radius 2.46 \(\times\) 10\(^4\) km and rotates once around its axis in about 16 h. (a) What is the gravitational force on a 3.00-kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

Calculate the earth's gravity force on a 75-kg astronaut who is repairing the Hubble Space Telescope 600 km above the earth's surface, and then compare this value with his weight at the earth's surface. In view of your result, explain why it is said that astronauts are weightless when they orbit the earth in a satellite such as a space shuttle. Is it because the gravitational pull of the earth is negligibly small?

Two stars, with masses \({M_1}\) and \({M_2}\), are in circular orbits around their center of mass. The star with mass \({M_1}\) has an orbit of radius \({R_1}\); the star with mass \({M_2}\) has an orbit of radius \({R_2}\). (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses\(-\)that is, \({R_1}\)/\({R_2}\) \(=\) \({M_2}\)/\({M_1}\). (b) Explain why the two stars have the same orbital period, and show that the period \(T\) is given by \(T = 2\pi\)(R1 + R2)\(^{3/2}\)/\(\sqrt{G(M1 + M2)}\). (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 km/s. The second star, Beta, has an orbital speed of 12.0 km/s. The orbital period is 137 d. What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.28). The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

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