21 worked answers · short steps first, explanations when you need them
Chapter 7 Exercises
Textbook pages 130–134
Question 1
(a) Can gravity be shielded like electric force inside a hollow conductor? (b) Could an astronaut detect gravity in a very large orbiting space station? (c) The Sun pulls Earth more strongly than the Moon does, yet the Moon’s tides are stronger. Why?
Steps only
(a) No known material or hollow sphere blocks gravity. A spherical shell’s own field is zero inside, but it does not block gravity from outside bodies. (b) Yes. Across a large station, gravity changes slightly from one end to the other, so the astronaut could detect a tidal effect.
Answer
(a) No; gravity cannot be shielded. (b) Yes, a large station can reveal a small gravity difference. (c) The Moon’s much shorter distance gives it the stronger tidal effect.
Question 2
Choose the correct word: (a) g increases/decreases with altitude; (b) g increases/decreases with depth in a uniform Earth; (c) g is independent of Earth’s mass/body’s mass; (d) −GMm(1/r₂−1/r₁) is more/less accurate than mg(r₂−r₁).
Steps only
As distance from Earth’s centre increases, g decreases. Inside a uniform Earth, g decreases with depth and is zero at the centre. The acceleration of a test body does not depend on its own mass.
Answer
(a) Decreases; (b) decreases; (c) body’s mass; (d) more accurate.
Question 3
A planet orbits the Sun with a period half of Earth’s. What is its orbital size compared with Earth’s?
Steps only
Kepler’s third law gives T²∝r³, so rplanet/rEarth=(Tplanet/TEarth)²ᐟ³.
Answer
About 0.63 times Earth’s orbital size.
Question 4
Jupiter’s moon Io has orbital period 1.769 days and orbital radius 4.22×10⁸ m. Use this to show Jupiter’s mass is about one-thousandth of the Sun’s.
Steps only
For a satellite of mass m orbiting a much heavier mass M, T²=4π²r³/(GM), so M=4π²r³/(GT²). Convert T=1.769×86400 s.
Answer
Jupiter’s mass is about 1.9×10²⁷ kg, roughly one-thousandth of the Sun’s.
Question 5
Model the Milky Way as 2.5×10¹¹ stars, each with one solar mass. How long does a star 50,000 light-years from the centre take to make one revolution? Use one light-year=9.46×10¹⁵ m and MSun=2.0×10³⁰ kg.
Steps only
Treat the galaxy’s mass as concentrated at its centre. For a circular orbit, T=2π√(r³/GM). Here r=50,000×9.46×10¹⁵=4.73×10²⁰ m and M=2.5×10¹¹×2.0×10³⁰=5.0×10⁴¹ kg.
Answer
About 3.6×10⁸ years (360 million years).
Question 6
(a) For a satellite with gravitational potential energy zero at infinity, its total energy is the negative of its kinetic/potential energy. (b) Is less or more energy needed to escape Earth from an orbit than from rest at the same height?
Steps only
For a circular orbit, U=−2K, so total energy E=K+U=−K.
Answer
(a) Kinetic energy; (b) less energy for the orbiting satellite.
Question 7
Does escape speed depend on (a) the projectile’s mass, (b) its launch location, (c) projection direction, (d) launch height?
Steps only
At distance r from Earth’s centre, escape speed is √(2GM/r). The projectile’s own mass cancels, and the direction does not affect the minimum speed if air resistance and other bodies are ignored.
Answer
(a) No; (b) only if the location changes the distance from Earth’s centre; (c) no; (d) yes.
Question 8
For a comet in an elliptical orbit, say which quantities stay constant: (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total mechanical energy. Ignore mass loss.
Steps only
Gravity is a central force, so the comet’s angular momentum about the Sun is conserved. Gravity is also conservative, so total mechanical energy is conserved.
Answer
Constant: angular momentum and total mechanical energy. Not constant: linear speed, angular speed, kinetic energy and potential energy.
Question 9
Which effects may an astronaut in orbit experience: (a) swollen feet, (b) swollen face, (c) headache, (d) difficulty with orientation?
Steps only
In weightlessness, body fluids shift toward the upper body, so the face may look swollen and the feet may feel less full. The unfamiliar free-fall environment can also cause headache and orientation difficulty.
Answer
(b), (c) and (d): swollen face, headache and orientation difficulty.
Question 10
At the centre of a uniform hemispherical shell, which way does gravitational intensity point in the diagram: arrow a, b, c, or zero?
Steps only
The sideways components from opposite parts of the hemisphere cancel. The remaining components add toward the curved shell, along the symmetry axis into the bowl.
Answer
Arrow c: along the symmetry axis toward the hemispherical shell.
Question 11
For the same hemispherical shell, which arrow d, e, f or g gives gravitational intensity at point P in the diagram?
Steps only
The field points toward the shell’s mass. At point P, symmetry cancels the sideways components; the net field points along the axis toward the denser curved part.
Answer
Arrow e: toward the shell along the symmetry direction.
Question 12
A rocket travels from Earth toward the Sun. At what distance from Earth’s centre is the net gravitational force zero? Use MEarth=6×10²⁴ kg, MSun=2×10³⁰ kg and separation D=1.5×10¹¹ m.
Steps only
At the zero-force point, the pulls are equal: MEarth/x²=MSun/(D−x)². Taking positive square roots gives (D−x)/x=√(MSun/MEarth).
Answer
About 2.6×10⁸ m from Earth’s centre toward the Sun.
Question 13
How can we estimate the Sun’s mass from Earth’s orbit? Earth’s mean orbital radius is 1.5×10⁸ km. Use Earth’s orbital period 1 year and G=6.67×10⁻¹¹ SI.
Steps only
For Earth’s nearly circular orbit, gravity supplies centripetal force: GM☉m/r²=mv²/r. With v=2πr/T, this becomes M☉=4π²r³/(GT²).
Answer
The Sun’s mass is about 2.0×10³⁰ kg.
Question 14
Saturn’s year is 29.5 Earth years. Earth is 1.50×10⁸ km from the Sun. Estimate Saturn’s orbital distance.
Steps only
Kepler’s third law gives rS/rE=(TS/TE)²ᐟ³.
Answer
About 1.43×10⁹ km from the Sun.
Question 15
A body weighs 63 N at Earth’s surface. What gravitational force acts on it at a height equal to half Earth’s radius?
Steps only
At height h=R/2, distance from Earth’s centre is 3R/2. Gravitational force varies as 1/r².
Answer
28 N.
Question 16
Earth is assumed to have uniform density. A body weighs 250 N at the surface. What would it weigh halfway down to Earth’s centre?
Steps only
Inside a uniform Earth, only the mass inside radius r contributes, and g is proportional to r. Halfway to the centre means r=R/2.
Answer
125 N.
Question 17
A rocket is fired vertically from Earth’s surface at 5 km/s. How high above Earth’s surface does it rise before returning? Use M=6.0×10²⁴ kg, R=6.4×10⁶ m and G=6.67×10⁻¹¹ SI.
Steps only
At launch and at the highest point, conserve energy: ½v²−GM/R=−GM/rmax, where rmax is measured from Earth’s centre.
Answer
It rises about 1.6×10⁶ m (1,600 km) above Earth’s surface.
Question 18
Earth’s escape speed is 11.2 km/s. A body is launched at three times this speed. Find its speed very far from Earth, ignoring the Sun and other planets.
Steps only
Use energy conservation with zero potential energy at infinity: ½v∞²=½v₀²−GM/R. Since ve²=2GM/R, v∞²=v₀²−ve².
Answer
About 31.7 km/s.
Question 19
A 200 kg satellite orbits 400 km above Earth. How much energy must be supplied to send it out of Earth’s gravitational influence? Use MEarth=6.0×10²⁴ kg, R=6.4×10⁶ m and G=6.67×10⁻¹¹ SI.
Steps only
Orbital radius r=R+h=6.8×10⁶ m. A circular satellite’s total energy is −GMm/(2r), so that much energy must be added to reach zero energy at infinity.
Answer
About 5.9×10⁹ J.
Question 20
Two equal stars, each of mass 2×10³⁰ kg, start nearly at rest 10⁹ km apart and fall directly toward each other. Each star has radius 10⁴ km. Find each star’s speed when they touch. Use G=6.67×10⁻¹¹ SI.
Steps only
At contact, centre separation is twice a star’s radius: 2×10⁷ m. By symmetry both stars have the same speed v. The loss of gravitational potential energy becomes their combined kinetic energy: GM²(1/rf−1/ri)=Mv².
Answer
Each star moves at about 2.6×10⁶ m/s when they touch.
Question 21
Two spheres, each mass 100 kg and radius 0.10 m, have centres 1.0 m apart on a horizontal table. Find the net gravitational force and gravitational potential at the midpoint. Is equilibrium there stable?
Steps only
The midpoint is 0.50 m from each centre. Equal attractions point in opposite directions, so net force is zero. Potential is a scalar and adds: Φ=−2GM/r.
Answer
Net force: 0 N. Potential: −2.67×10⁻⁸ J/kg. The equilibrium is unstable.