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Chapter 13: Problem 44

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think we should land a vehicle there to search for life. Before launching it, we would want to test such a lander under the gravity conditions at the surface of Europa. One way to do this is to put the lander at the end of a rotating arm in an orbiting earth satellite. If the arm is 4.25 m long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the lander is the same as the acceleration due to gravity at the surface of Europa? The mass of Europa is 4.80 \(\times\) 10\(^{22}\) kg and its diameter is 3120 km.

Short Answer

Expert verified
The required angular speed is approximately 5.31 rpm.

Step by step solution

01

Calculate Europa's Surface Gravity

The formula for gravitational acceleration is given by \[ g = \frac{G \cdot M}{R^2} \]where:- \( G = 6.674 \times 10^{-11} \text{ m}^3 / (\text{kg} \cdot \text{s}^2) \) (the universal gravitational constant),- \( M = 4.80 \times 10^{22} \text{ kg} \) (mass of Europa), and - \( R \) is the radius of Europa, which is half of its diameter: \( 1560 \times 10^3 \text{ m} \).Plug in these values to compute \( g \): \[ g_E = \frac{6.674 \times 10^{-11} \times 4.80 \times 10^{22}}{(1560 \times 10^3)^2} \approx 1.314 \text{ m/s}^2 \].
02

Relate Centripetal Acceleration to Angular Speed

To achieve the same acceleration using a rotating arm, the centripetal acceleration formula is used:\[ a_c = r \cdot \omega^2 \]where:- \( a_c = g_E = 1.314 \text{ m/s}^2 \) (given from Step 1)- \( r = 4.25 \text{ m} \) (length of the arm)- \( \omega \) is the angular speed in radians per second.We need to find \( \omega \). Rearrange:\[ \omega^2 = \frac{a_c}{r} \text{, so}\]\[ \omega = \sqrt{\frac{1.314}{4.25}} \].
03

Calculate Angular Speed in Radians per Second

Continue from Step 2:\[ \omega = \sqrt{\frac{1.314}{4.25}} \approx 0.556 \text{ rad/s} \].
04

Convert Angular Speed to Revolutions per Minute (rpm)

To convert from radians per second to revolutions per minute (rpm), use the conversion factors:1. 1 revolution = \( 2\pi \) radians,2. 1 minute = 60 seconds.Thus:\[ \text{rpm} = \omega \times \frac{1 \text{ rev}}{2 \pi \text{ rad}} \times \frac{60 \text{ s}}{1 \text{ min}} \approx 0.556 \times \frac{60}{2\pi} \approx 5.31 \text{ rpm} \].

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is an essential concept in physics whenever an object is in circular motion, like a car turning around a bend or a lander being swung by an arm. This type of acceleration always points towards the center of the circle along which the object is moving. It is what keeps the object in its circular path instead of flying off in a straight line. To calculate centripetal acceleration, we use the formula:\[ a_c = r \cdot \omega^2 \]where:- \( a_c \) is the centripetal acceleration,- \( r \) is the radius of the circular path, and - \( \omega \) is the angular speed in radians per second.The centripetal acceleration depends on how fast the object is spinning (speed) and how far away it is from the center (radius). The faster the spin or the longer the arm, the greater the acceleration experienced.
Universal Gravitational Constant
The Universal Gravitational Constant, denoted as \( G \), is a key component in the law of universal gravitation, developed by Sir Isaac Newton. This constant characterizes the strength of gravitational attraction between two bodies. Its value is always the same: \[ G = 6.674 \times 10^{-11} \text{ m}^3 / (\text{kg} \cdot \text{s}^2) \]It allows us to calculate the gravitational force between two masses. In the context of finding surface gravity like on Europa, \( G \) helps to determine the gravitational pull by: \[ F = \frac{G \cdot M_1 \cdot M_2}{r^2} \]where \( M_1 \) and \( M_2 \) are the masses of the two objects and \( r \) is the distance between their centers. This formula shows that gravitational force decreases with the square of the distance, making \( G \) an essential tool for understanding gravitational interactions in the universe.
Europa's Surface Gravity
Europa's surface gravity is the acceleration due to gravity experienced by objects on its surface. It is crucial to understand when planning missions, as this gravity determines how objects like spacecraft will behave when they land. To find this, we employ the formula for gravitational acceleration:\[ g = \frac{G \cdot M}{R^2} \]where:- \( G \) is the universal gravitational constant,- \( M \) is the mass of Europa, and- \( R \) is the radius of Europa.Plugging in the values as provided, Europa's surface gravity \( g_E \) is calculated to be about \( 1.314 \text{ m/s}^2 \). This is significantly weaker than Earth's gravity, which is roughly \( 9.81 \text{ m/s}^2 \). Knowing Europa's surface gravity helps engineers design spacecraft that can maneuver and land safely on the icy moon, even under these reduced gravitational forces.

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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?

A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. One of them has a mass of 65 kg and the other a mass of 72 kg, and they start from rest 20.0 m apart. (a) Make a free- body diagram of each astronaut, and use it to find his or her initial acceleration. As a rough approximation, we can model the astronauts as uniform spheres. (b) If the astronauts' acceleration remained constant, how many days would they have to wait before reaching each other? (Careful! They \(both\) have acceleration toward each other.) (c) Would their acceleration, in fact, remain constant? If not, would it increase or decrease? Why?

A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_E\) of the earth. Ignoring air resistance, derive an expression for the speed y of the hammer when it reaches the earth's surface. Your expression should involve \(h\), \(R_E\), and \(m_E\) (the earth's mass).

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{1700}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

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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