Solution Manual of Conceptual Physics 12E – Hewitt – SM
9-1. (a) d = ? Comparing the two gravitational fields, Since the gravitational field, like the gravitational force, is inversely proportional to the square of the distance if g is 100 times smaller, then d must be 10 times greater More formally, since we can write the proportion
(b) d = ? The distance from Earth’s center is 10RE. The distance from the surface is simply
9-2. (a) The mass is still m. Mass is independent of location.
9-3. (a) Fnew = ? The force is quadrupled.
(b) Fnew = ? The force is unchanged.
(c) Fnew = ? The force is halved.
9-4. (a) Fnew = ? This is the essence of an inverse-square law. When the distance changes by k, the force changes by
9-5. (a) F = ? Each ball has mass m. The distance between centers is 2R. So the force needed to separate them is the same as the attractive force between them =
(c) The result tells us that the gravitation force between two objects is negligible compared to their weight at Earth’s surface.
9-6. (a) Fnew = ? From quadrupling the mass of the star quadruples the force at its surface. Doubling the diameter doubles the center-to-surface distance, which reduces the force at the surface by a factor of four. Taken together, the effects of these two changes cancel out, leaving the field unchanged overall. More formally: At the star’s surface The force is unchanged.
(b) Fnew = ? Let’s call R the radius of the star. Then from reducing R by a factor of 10 should increase g by a factor of 100. More formally: The field is 100 times stronger.
9-7. (a) Assume that we can model you and the charming someone as point particles separated by a distance x. Then
(b) From Newton’s third law, the charming person’s attraction to you is just as much as your attraction to them .
(c) (an incredibly small force, about the weight of a speck of dust).
9-8. (a) gP = ? Since the hanging mass is in equilibrium, the upward force from the spring (kx) equals the downward force of gravity (mg) in both cases.
(b) From if the radius of the planet were to increase, the force on the hanging mass would decrease and the spring would be stretched less.
9-9. (a) d = ? Since we are comparing the force at two different distances, this suggests that we consider the ratio of the two. Since At twice the distance the force will be one-fourth as much.
(b) From above
9-10. (a) dfrom Earth= ? We’re looking for the spot where Fnet = 0. That is, where Canceling out like factors and taking the square root of both sides gives
(b) from Earth’s center. The spot is from the center of the Moon.
9-11. (a) As always, the acceleration of a body due to gravity is independent of the mass of the body.
(d) Clearly there is gravity in space! That is what keeps the ISS (and the Moon, for that matter) in orbit around Earth. From (c), the gravitational pull on an astronaut in the ISS is 90% of the pull on the surface. Definitely not zero!
(e) If you were on the ISS you’d need to throw the coin “backward” at a velocity exactly equal and opposite to the habitat’s velocity. Then the coin would have zero horizontal velocity relative to Earth and fall in a straight line to Earth. *
9-12. (a) gwhite dwarf = ? From So we see that gwhite dwarf is much greater, bigger than gE by the factor That is,
(c) If the radius were halved again g would be four times greater than in (a) .
(b) h = ? Calling upward the positive direction,
(c) t = ? Calling upward the positive direction,
9-14. (a) MP = ? The scale reading is the normal force on you, equal to the gravitational attraction between you and the planet.
(b) Half the diameter means half the radius, which means four times the force:
(c) Your weight would be zero. No mass pulls inward on you. You would be surrounded by a symmetric distribution of matter, and the vector sum of the gravitational force on you from each bit of that symmetric distribution surrounding you cancels to zero.
9-17. (a) The Sun’s pull on the blob is 177 times stronger harder.
(b) Tides have to do with the difference in pulls on the opposite sides of Earth. The difference between the Moon’s pull on the near side of Earth and the far side of Earth is greater than the same difference due to the pull of the Sun.
9-18. Pull = ? By mass, the Sun pulls 330,000 times harder on you than Earth does. By distance the Sun pulls as hard on you. So So Earth pulls harder on you. More formally, Taking the ratio of these two forces: Again, Earth pulls harder on you.
9-19. (a) From Two Earth radii above Earth’s surface is three Earth radii from Earth’s center. At three times the distance from Earth’s center, g is one-ninth as strong so the distance risen is nine times greater.
9-20. (a) Add the three forces on one asteroid due to the others. The diagram considers the forces on the upper-left asteroid:
Force Due To: x-component y-component
Lower left mass
Upper right mass
Lower right mass cos45º
SUM TOTAL (1 + )
–(1 + )
The force will be the same on any of the masses, and the direction will always be pointing diagonally across the square from the selected mass.
(b) The force on a dust speck at the center will be zero. Since the masses are distributed symmetrically about the center, the attraction from each pair of opposing masses will cancel out.
9-22. As acceleration due to gravity, g has unit of acceleration, As gravitational field strength, g is , which has units . The units are equivalent:
9-23. For a freely falling body of mass m, For a freely-falling body of mass 2m,
9-24. Assume that we can model the twins as uniform spheres with centers separated by 1 m. Let mT be the mass of each twin. The force between the twins is The weight of the dust speck is These two forces are about the same.
9-28. The questions asks us to compare the gravitational force at two different locations, which suggests that we set up the problem as a ratio:
9-29. (a) gVenus = ?
9-31. (a) To decrease g to 0.0001 m/s2 (the value of g on the asteroid’s surface), g has to decrease by a factor So d has to increase by a factor of . So you have to be about 300 Earth radii farther from Earth’s center to experience this same acceleration due to gravity.
9-32. (a) (W = mg produces essentially the same answer).
(b) Notice that the distance we use is the distance from the center of Earth, REarth + h, and that all distances have to be in meters to be consistent with the units of G.
9-33. rJack = ? We want to find rJack such that . Since d must increase to of its original value, More formally:
9-34. The Sun pulls with more than twice as much force on the Moon as Earth does.
9-36. (a) We want to calculate the gravitational force between the Sun and 1 kg of water where the water is on the side of Earth nearer the Sun, a distance dES – RE away from the Sun; calculate the force when the 1 kg of water is on the side of Earth farther from the Sun, a distance dES + RE away from the Sun; and then take the difference of these two forces. Formally: =1.00 10-6 N.
(b) Likewise, here we want to calculate the gravitational force between the Moon and 1 kg of water where the water is on the side of Earth nearer the Moon, a distance dEM – RE away from the Moon; calculate the force when the 1 kg of water on the side of Earth farther from the Moon, a distance dEM + RE away from the Moon; and then take the difference of these two forces. Formally: =2.21 10-6 N.
(c) Tides have to do with the difference in pulls on the opposite sides of Earth. The effect of the Moon’s gravity on Earth’s tides is 2.2 times larger than that of the Sun’s gravity.
9-37. Located two Earth radii above the Earth, the melon is three Earth radii from Earth’s center.
The force from the mountain pulling on you is larger than the force on you from the Moon.
9-39. To simplify, consider the mom and her bouncing baby to be spheres. Then The force of the mom on the baby is more than eight times greater than the force of Mars on the baby.