Ch 13.1 #5 Three uniform spheres of mass 2.00 kg, 4.00 kg, and 6.00 kg are placed at the corners of a right triangle. Calculate the resultant gravitational force on the 4.00 kg object.
	The 4 kg object has two objects attracting it, thus you need to add both objects individually. F = GMm/r^2
	F6:4 = -6.673 x 10^-11 6(4)/(4^2) i F6:4 = -10.0 x 10^-11 i Newtons	F2:4 = 6.673 x 10^-11 6(4)/(4^2) j F2:4 = 5.93 x 10^-11 j Newtons
	F2,6:4 = (-10.0 x 10^-11 i + 5.93 x 10^-11 j ) Newtons

Ch 13.2 Gliese 581 C recently has marked a milestone in the search for worlds beyond our solar system. It is the smallest exoplanet ever detected, and the first to lie within the habitable zone of its parent star, thus raising the possibility that its surface could sustain liquid water, or even life. It is 50 percent bigger and five times more massive than Earth.

(a) What is the gravity on the surface of this planet? Ans: 21.8 m/s^2 (b) A student at Gliese College has been given a project to verify the value of “G”. This student suspends two 50.0 meter wires from a tall ceiling that are 2.00 meters apart. At the bottom of the wires this student has attached Iridium (ρIr = 22.2 g/cc) balls (each R = 0.900 meter), which are deflected toward each other by how much? Ans: each one by 2.6 μm

(a) GMEm/rE^2 / gE = GMGlim/rGli^2 / gGli ME/rE^2 / gE = MGli/rGli^2 / gGli ME/rE^2 / 9.81 = 5ME/1.5rE^2 / gGli gGliese = 21.8 m/s^2		at small angles tanθ ≈ sinθ ≈ θ	(b) Distance between objects are 2 m or (2 – 2x) ≈ 2 because x << 2m
			tanθ = x/50 θ = x / 50	tanθ = Fg / mg θ = GM/ r^2 / g θ = G 67,700/ g(2-2x)^2
V = 4πR^3/3 V = 4π 0.9^3/3 V = 3.05	m = ρ V m = 22,200 (3.05) m = 67,700 kg		x / 50 = G 67,700 / 21.8*2^2 r = 2.59 x 10^-6 meters for each ball

Ch 13.3 #10

The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. If the radius of the Moon is about 0.250 R_E, find the ratio of their average densities ρ_moon / ρ_earth . ans 2/3

mg = GMm / r^2 g = G M / r^2 g = G ρV / r^2 g = G ρ(4/3 πr^3) / r^2 g = 4/3 Gρπr

gmoon / gearth = 4/3 Gρπrmoon / 4/3 Gρπrearth gmoon / gearth = 1 / 6 1 / 6 = 4/3 Gρπrm / 4/3 Gρπre 1 / 6 = ρmrm / ρere ρere = 6ρm rm ρere = 6ρm0.25re 2/3ρe = ρm

Ch 13.4 #20

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose that the mass of a certain spherical neutron star is twice the mass of the sun and its radius is 10.0 km. Determine the greatest possible angular speed it can have so that the matter at the surface of the star on its equator is just held in orbit by the gravitational force.

FNeutron:matter = G MN m / r^2	Fc = mv^2/r	Show Algebra G MN m / r^2= mv^2/r G MN / r= v^2 (ωr)^2 = G MN / r ω = (G M2xSun / r^3)^1/2
FN:m = Fc G MN m / r^2= mv^2/r ω = (6.673x10^-11 (2*1.991x10^30) / (10^4)^3)^1/2 ω = 16,300 radians per second

Ch 13.5 #25 Compute the magnitude and direction of the gravitational field at a point P on the perpendicular bisector of the line joining two objects of equal mass separated by a distance 2a. ans: g = 2MGr((a^2 + r^2)^-3/2

We know the y components are equal and opposite and that both x components are pulling at Point P in the –x direction.
gy = g1y – g2y = 0 g = g1x + g2x g1x = g2x g = 2g1x (-i )	g1x = cosθ g1 g1x = r/(a^2+r^2)^1/2 GM/(a^2 + r^2) g1x = GMr/(a^2+r^2)^3/2 g = 2 g1x (-i ) g = 2(GMr/(a^2+r^2)^3/2) (-i )	cos θ = adj / hyp cos θ = r / (a^2 + r^2)^1/2 mg1 = GMm/r^2 g1 = GM/(a^2 + r^2)

Ch 13.6 #31 A system consists of three particles, each of mass 5.00 g, located at corners of an equilateral triangle with sides of 30.0 cm. (a) Calculate the potential energy of the system (b) If the particles are released simultaneously, where will they collide? ans: (a) -1.67 x 10^-14 Joules, (b) at the center

Utotal = U1:2 + U1:3 + U2:3 U1:2 = U1:3 = U2:3 Utotal = 3 U1:2

U1:2 = G M m / r U1:2 = 6.673x10^-11(0.005)(0.005)/ 0.3 U1:2 = 5.56 x 10^-15

Utotal = 3 U1:2 Utotal = 3 * 5.56 x 10^-15 Utotal = 16.7 x 10^-15 Joules

Ch 13.7 #40

The planet Uranus has a mass about 14 times the Earth’s mass, and its radius is equal to about 3.7 Earth radii. (a) By setting up ratios with the corresponding Earth values, find the free-fall acceleration at the cloud tops of Uranus. (b) By ignoring the rotation of the planet, find the minimum escape speed from Uranus.

(a) mg = GMm/r^2 9.8 = G ME / rE^2 gUr = G 14ME / (3.7rE)^2 gUr = 9.8(14 / (3.7)^2) gUr =10.0 m/s^2

(b) vesc,E = (G ME / rE)^1/2 vesc,Ur = (G 14ME / 3.7rE)^1/2 11200 = 1 vesc,Ur = (14 / 3.7)^1/2 vesc,Ur = 11,200 (14 / 3.7)^1/2 vesc,Ur = 2.18 x 10^4 m/s

vesc = √(2GM/RE) vesc,E = 1.12 x 10^4 m/s