Question

b 6. two artificial satellites, 1 and 2, orbit the earth in circular orbits having radii r₁ and r₂, respectively, as shown above. if r₂ = 2r₁, the accelerations a₂ and a₁ of the two satellites are related by which of the following? (a) a₂ = 4a₁ (b) a₂ = 2a₁ (c) a₂ = a₁ (d) a₂ = a₁/2 (e) a₂ = a₁/4 e 7. two identical stars, a fixed distance d apart, revolve in a circle about their mutual center of mass, as shown above. each star has mass m and speed v. g is the universal gravitational constant. which of the following is a correct relationship among these quantities? (a) v² = gm/d (b) v² = gm/2d (c) v² = gm/d² (d) v² = mgd (e) v² = 2gm²/d

Explanation:

Problem 6:

Step1: Use centripetal - force and gravitational - force relation

For a satellite of mass \(m\) orbiting the Earth (mass \(M_E\)) in a circular orbit of radius \(R\), the centripetal force \(F_c = ma\) is provided by the gravitational force \(F_g=\frac{GM_Em}{R^{2}}\). So, \(ma=\frac{GM_Em}{R^{2}}\), and the acceleration \(a = \frac{GM_E}{R^{2}}\) (where \(G\) is the gravitational constant and \(M_E\) is the mass of the Earth).

Step2: Find the ratio of accelerations

For satellite 1, \(a_1=\frac{GM_E}{R_1^{2}}\), and for satellite 2, \(a_2=\frac{GM_E}{R_2^{2}}\). Given \(R_2 = 2R_1\), then \(a_2=\frac{GM_E}{(2R_1)^{2}}=\frac{GM_E}{4R_1^{2}}\). Substituting \(a_1=\frac{GM_E}{R_1^{2}}\) into the expression for \(a_2\), we get \(a_2=\frac{a_1}{4}\).

Step1: Calculate the distance from each star to the center of mass

The two - star system: since the stars are identical (\(M_1 = M_2 = M\)), the distance from each star to the center of mass \(r=\frac{D}{2}\).

Step2: Equate the gravitational force and centripetal force

The gravitational force between the two stars \(F_g=\frac{GM\cdot M}{D^{2}}\). The centripetal force on one star \(F_c = M\frac{v^{2}}{r}\). Substituting \(r = \frac{D}{2}\) into the centripetal - force formula, we have \(F_c = M\frac{2v^{2}}{D}\).

Step3: Set \(F_g=F_c\)

\(\frac{GM\cdot M}{D^{2}}=M\frac{2v^{2}}{D}\). Cancel out one \(M\) from both sides of the equation: \(\frac{GM}{D^{2}}=\frac{2v^{2}}{D}\). Then \(v^{2}=\frac{GM}{2D}\).

Answer:

E. \(a_2=a_1/4\)

Problem 7: