PH 2213 : Chapter 06 Homework Solutions Problem 6.08 : Every few hundred years most of the planets line up on the same side of the Sun. Calculate the total force on the Earth due to Venus, Jupiter, and Saturn, assuming all four planets are in a line. The masses are M ven = 0.815M e, M jup = 318M e M sat = 95.1M e and the mean distances of the four planets from the Sun are 108, 150, 778, and 1430 million km respectively. What fraction of the Sun s force on the Earth is this? Coordinate system: let s say that positive is radially outward from the Sun. Then Venus will be exerting a force on the Earth in the negative direction, the other two planets will be exerting a force on the Earth in the positive direction. The masses of the other planets were given in terms of the mass of the Earth, which we find in the tables at the front of the book to be: M e = 5.98 10 24 kg. For each of the three planets, the magnitude of the force will be F = Gm 1 m 2 /r 2 where m 1 and m 2 are the masses of the corresponding planets, and r is the distance between them. Venus orbits the Sun at 108 million km and the Earth orbits the Sun at 150 million km, so in the configuration described, the distance between the Earth and Venus is 150 108 = 42 million km or 42 10 6 10 3 meters or finally r = 4.2 10 10 m. The force between Venus and Earth then will be: F = (6.67 10 11 ) (0.815)(5.98 1024 )(5.98 10 24 ) (4.2 10 10 ) 2 = 1.102 10 18 N. Jupiter : doing the same process as above with Jupiter being 318 times the mass of the earth, and a separation distance of 778 150 = 628 million km or 6.28 10 11 m we have F = 1.923 10 18 N. Saturn : same with a mass of 95.1 times that of the Earth and a separation distance of 1430 150 = 1280 million km or 1.28 10 12 m we have F = 0.138 10 18 N. Total Force : careful with signs now. Venus force is in our negative direction, the other two are in the positive direction so ΣF = ( 1.102 + 1.923 + 0.138) 10 18 = 0.959 10 18 or 9.59 10 17 N. (The positive direction of the coordinate system wasn t specified in the problem, but the homework system accepted plus or minus this number.) Sun : That sounds like a lot but how does this compare to the force the Sun is exerting on the Earth? F sun = (6.67 10 11 ) (5.98 1024 )(1.99 10 30 ) (150 10 9 ) 2 = 3.53 10 22 N. The problem asked us to compare these, so the sum of the forces of those other planets divided by the force from the sun is: ratio = (9.59 10 17 )/(3.53 10 22 ) = 2.72 10 5 (the answer they were looking for) or inverting that - the force of the Sun is about 37,000 times that of all those other planets combined (so even when those planets lined up, the effect on the Earth s orbit was negligible).
Problem 6.09 : Four 8.5 kg spheres are located at the corners of a square of side 0.8 m. Calculate the magnitude and direction of the gravitational force exerted on one sphere by the other three. The masses are all identical and are located at the corners of a perfect square, so we have symmetry here. Each mass will feel some net force directed towards the center of the square. Let s find the force that the other three masses are exerting on the mass located in the lower left corner. Force is a vector so we ll look at the x and y components separately, but from symmetry, the numerical value of the x and y components will be the same, so we really only need to compute one of the sums. Looking in the x (left-right) direction, the object on the lower right will be attracting the mass at the origin towards it with a force of F x = Gmm/d 2. The object on the upper left will have a vector force entirely in the Y direction, so won t have any X component. The object on the upper right is located at r = 2d away from the object at the origin, and that force vector is at a 45 o angle so it s X component will be F x = (Gmm/r 2 ) cos (45 o ) or F x = G mm ( 2 1. 2d) 2 The total force in the X direction then is: ΣF x = G m2 d 2 ΣF x = G m2 d 2 (1 + 1 2 2 ) + G m2 ( 2d) 2 1 2 and collecting terms: Using the numerical values we have here, with m = 8.5 kg and d = 0.8 m, we have ΣF x = 1.019 10 8 N. Similarly, ΣF y = 1.019 10 8 N. We now have the x and y components of the total force acting on the mass at the origin. The problem just wanted the magnitude of this force, so F = F 2 x + F 2 y = 1.44 10 8 N.
Problem 6.12 : Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9 10 22 kg and making the assumption that its density is the same as Earth s. An object of mass m on the surface of a planet or moon of mass M and radius R will feel a force of gravity of F G = GMm/R 2 which we can write as F G = mg with g = GM/R 2, as we discussed in class. Plugging in the mass and radius of the Earth at the front of the book, we arrive at g = 9.8 m/s 2, for example. Here we have the mass of Europa, but not it s radius but we can get that if we assume the density (ρ) is the same as the Earth s ρ = (density) = (mass)/(volume) and we know the mass and size of the Earth so could compute it s average density (which turns out to be about 5500 kg/m 3 ), then use that density and the given mass of Europa to find it s volume (and then from that, determine it s radius). That s a lot of steps though, so let s take a shortcut here: ρ = (density) = (mass)/(volume) so if Europa has the same density as the earth, it s ratio of mass to volume is the same as the Earth s: ρ = Meuropa V europa = M earth V earth V europa = V earth M europa M earth. or rearranging: The volume of a sphere is proportional to the cube of the radius though (V = 4 3 πr3 ), so this ratio can be re-written as: Reuropa 3 = Rearth 3 and taking the cube root of each side: M europa M earth R europa = R earth ( Meuropa M earth ) (1/3) Using the radius and mass of the Earth given at the front of the book: R europa = (6380 km)( 4.9 1022 5.98 10 24 ) 1/3 which yields R europa = (6380 km)(0.2016) = 1286.2 km or R = 1.2862 10 6 m which is only slightly smaller than the Earth s moon. (Note: the true radius of Europa is about 1550 km which is higher than what we found because it s density is actually a bit less than the Earth s density...) The value of g at the surface of Europa then would be: g = GM/R 2 = (6.67 10 11 )(4.9 10 22 )/(1286200) 2 = 1.975 m/s 2. That s about 1/5th of g here on the surface of the Earth, and slightly higher than g on the surface of the Moon, even though Europa is slightly smaller than the moon, but this result is due to the (erroneous) assumption that Europa had the same density as the Earth. We know the density of our moon is considerably lower than the Earth s average density, so should expect Europa to be different also, which it is. We know the mass and size of Europa pretty accurately now, and it s mean density is only about 3000 kg/m 3 (compared to the Earth s 5500 kg/m 3 ). The actual acceleration due to gravity on Europa s surface is about 1.3 m/s 2.
Problem 6.24 : Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 5800 km. This satellite is orbiting at that height above the surface of the Earth, which has a radius of 6380 km, so the circle this satellite is moving in has a radius of 6380 + 5800 = 12, 180 km or r = 1.218 10 7 m. Here we have a fairly tiny mass moving in a circle about a much more massive one. The satellite is undergoing a radial acceleration of a = v 2 /r which means it must be feeling a radial force of F = ma = mv 2 /r. This force is being provided by the gravitational attraction between the Earth (M) and the satellite m. F = ma here then becomes: GMm/r 2 = mv 2 /r which, after some cancellation and rearranging, yields a general relationship between the orbit speed v and the radius: v = GM/r Here the central body is the Earth with M = 5.98 10 24 kg and we have the radius of the circular orbit being r = 1.218 10 7 m so plugging in those (and the known value for G = 6.67 10 11 we find that v = 5723 m/s (about 12, 800 miles/hr). What would the orbital period of this satellite be? T = 2πr/v so T = (2)(π)(1.218 10 7 m)/(5723 m/s) = 13, 372 sec or about 3.7 hours.
Problem 6.56 : During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of about 100 km. How long did it take to go around the Moon once? We did a problem like this in class, looking at the orbital period of a satellite in near-earth orbit. Here the central object is the Moon instead of the Earth. For a light object in a circular orbit around a much heavier object, the period is related to the radius r of the circle and the mass M of the central object: T 2 = 4π2 GM r3 The radius r is the radius of the circle represented by the object s orbit. The command module is orbiting 100 km above the surface of the Moon, so the radius of the orbit will be the radius of the moon, plus 100 km. Looking at the inside front cover of the book, we find that the radius of the moon is 1740 km, so here r = 1740 + 100 = 1840 km = 1, 840, 000 m. We also find the mass of the moon there: M = 7.35 10 22 kg. Everything is in standard metric units now, so we can apply the equation to determine the period (which will come out in seconds): T 2 = 4π2 GM r3 so T 2 = 4π 2 (6.67 10 11 )(7.35 10 22 ) (1, 840, 000)3 = 5.016 10 7 so T = 5.016 10 7 = 7083 s (about 118 minutes, or nearly 2 hours). Compare that to the orbital period we got for satellites in near-earth orbit, which was about 90 minutes. Even though the object orbiting the moon was making a much smaller circle, the mass of the moon is so much lighter that the overall effect is it takes longer for an object orbiting just above the moon s surface to go once around than it takes for an object orbiting just above the earth s surface to go once around. We can push this a little further. Let s look at the general problem of a satellite orbiting around a moon or planet (or sun) where it s nearly skimming along the surface of the central body. That was nearly the case for this problem, and the one we did in class when we talked about near-earth orbiting satellites. Density is defined as mass per volume: ρ = M/V. For a spherical object, V = 4 3 πr3 and we can combine these to find that r 3 = 3M. Substituting that expression for 4πρ r3 into T 2 = 4π2 GM r3 we end up with T = (3π)/(Gρ). That implies the period for something nearly skimming the surface of another object only depends on that object s average density. If the moon and earth had the same density, satellites skimming their surfaces would have the same period, even though the earth is much larger. If we know the period of something in low-orbit, we know the density of the object it s orbitting directly.
Problem 6.69 : A science fiction tale describes an artificial planet in the form of a band completely encircling a sun. The inhabitants live on the inside surface (where it is always noon). Imagine that this sun is exactly like our own, that the distance to the band is the same as th Earth-Sun distance (to make the climate temperate) and that the ring rotates quickly enough to produce an apparent gravity of g as on Earth. What will be the period of revolution (this planet s year) in Earth days? If we focus on a person standing on the inner surface of this ring, they are moving in a circle so will feel a radial acceleration of a r = v 2 /r and the goal here is to have the rotation be fast enough that a r = g = 9.8 m/s 2. We were given that the ring has a radius equal to the distance from the Sun to the Earth. From the inside cover of the book: r = 149.6 10 6 km or converting units, r = 1.496 10 11 m. a r = v 2 /r so f = ra r = (1.496 10 11 )(9.8) = 1.211 10 6 m/s. This would be the speed of a point on the ring as it rotates around the Sun (and is about 2.7 million miles per hour). How long will it take the ring to make one complete revolution? v = 2πr/T so T = 2πr/v = (2)(π)(1.496 10 11 )/(1.211 10 6 ) = 776, 189 seconds. Converting to days: T = 776189 s 1 min 1 hr 1 day 60 s 60 min 24 hr = 8.98 days. Note that a planet just in orbit around the Sun at this distance should take exactly 1 year to make one complete revolution, so this ring world would need to be spun up to rotate much more rapidly, to the point where it makes one rotation in just under 9 days. A series of books (the Ringworld series, by Larry Niven) takes place on a planet like this, and there are huge thrusters mounted along the ring to accelerate it to this rate. The stresses on the material making up the ring would be huge (and the author made up a new material to account for it), but carbon nanotubes are capable of undergoing such stresses, and have been considered for the (almost equally crazy) idea of building a space elevator to hoist objects into near Earth orbits without requiring any rocket propulsion.