0.1
In introductory physics laboratories, a typical Cavendish balance for measuring the gravi-tational constant G uses lead spheres with masses of 2.10 kg and 21.0 g whose centers are separated by about 3.90 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
The gravitational force between the two masses is:
F = GmM r2 = 6.67 × 10−11×2.10 × 21 × 10 −3 (3.90 × 10−2)2 = 1.93 × 10−3N
Miranda, a satellite of Uranus, is shown in part a of the figure below. It can be modeled as a sphere of radius 242 km and mass 6.68 ×1019 kg.
• a) Find the free-fall acceleration on its surface.
• b) A cliff on Miranda is 5.00 km high. It appears on the limb at the 11 o’clock position in part a of the figure above and is magnified in part b of the figure above. A devotee of extreme sports runs horizontally off the top of the cliff at 7.70 m/s. For what time interval is he in flight?
• c) How far from the base of the vertical cliff does he strike the icy surface of Miranda? • d) What is his vector impact velocity?
a)
The free-fall acceleration on Miranda’s surface can be derived by equating the gravitational force F = GmMr2 and the free-fall force mg :
mg = GmM r2 g = GM r2 = 6.67 × 10−11× 6.68 × 10 19 (242 × 103)2 = 0.0761 m/s2
b)
We can use the free-fall equation of motion under the above-calculated accel-eration ∆h =12gt2 to get the time in flight:
t = s 2∆h g = r 2 × 5000 0.0761 = 363 s
c)
How far from the base cliff will he strike can be evaluated by looking at the horizontal component of the equations, x = vxt:
x = 7.70 × 363 = 2791 m
d)
The horizontal component of his velocity is being constant throughout the motion vx = 7.70m/s, we can evaluate the vertical component at the impact
by using the time-independent equation: v2y− v 2 y0 = 2g∆h vy = p 2g∆h = √2 × 0.0761 × 5000 = 27.59 m/s and v = p27.592+ 7.702 = 28.6 m/s
The direction of his impact is :
θ = arctanvy vx
= arctan27.59 7.7 = 74.4◦
A comet (see figure below) approaches the Sun to within 0.570 AU, and its orbital period is 90.6 years. (AU is the symbol for astronomical unit, where 1 AU = 1.50 ×1011 m is the
mean EarthSun distance.) How far from the Sun will the comet travel before it starts its return journey.
Kepler’s Law relates the square of the orbital period of a planet to the cube of the semi-major axis (distance a in the figure), the proportionality constant is GM 4π2: a3 = GM 4π2T 2 = 6.76 × 10 −11× 1.989 × 1030 4π2 T 2 = 3.360 × 1018T2 = 3.360 × 1018× (90.6 × 365 × 24 × 3600)2 = 2.75833 × 1037m3
and a = 3.02145 × 1012m = 3.02145 × 10 12 1.50 × 1011 = 20.143 AU x = 2a − 0.570 = 39.71 AU
0.4
Neutron stars are extremely dense objects formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 11.0 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.
The matter at the surface of the neutron stars is subject to the gravitational force GMnm
r2 which is balanced by the centripetal
mv2
r force that keeps the star
in its orbit: GMnm r2 = mv2 r = m(rω) 2 r ω = r GMn r3 = r 6.76 × 10−11×2 × 1.989 × 1030 110003 = 14122 rad/s
How much work is done by the Moon’s gravitational field as a 995 kg meteor comes in from outer space and impacts on the Moon’s surface?
The work done by the Moon’s gravitational field can be evaluated through the change in potential energy:
W = −∆U = −(−GM m R − GM m ∞ ) = −(−GM m R − 0) = 6.67 × 10 −11× 7.36 × 1022× 995 1.74 × 106 = 2.81 × 109J
0.6
After the Sun exhausts its nuclear fuel, its ultimate fate will be to collapse to a white dwarf state. In this state, it would have approximately the same mass as it has now, but its radius would be equal to the radius of the Earth.
• a) Calculate the average density of the white dwarf. • b) Calculate the surface free-fall acceleration.
• c) Calculate the gravitational potential energy associated with a 4.93-kg object at the surface of the white dwarf.
a)
The white dwarf will have a mass almost the same mass as the sun Ms= 1.989×
The density is thus: ρ = 4πrMs3 3 = 1.989 × 10 30 4π×(6.37×106)3 3 = 1.84 × 109kg/m3
b)
Following the same steps in problem 2-a), the free fall acceleration is : g = GM r2 = 6.67 × 10−11× 1.989 × 10 30 (6.37 × 106)2 = 3.27 × 106m/s2
c)
Potential energy of the 4.93-kg object at the surface of the white dwarf is : Ug = − GMsm r = −6.67 × 10−11×1.989 × 10 30× 4.93 6.37 × 106 = −1.03 × 1014J
0.7
• a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth’s orbit?
• b) Voyager 1 achieved a maximum speed of 125,000 km/h on its way to photograph Jupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar system?
The value of −GM m
r is the potential energy that keeps the spacecraft in the
solar system and to overcome this barrier, the spacecraft will need a minimum kinetic energy 1
2mv
v = r = r 2 × 6.67 × 10−11× 1.989 × 1030 1.50 × 1011 = 42.1 km/s
b)
Again the same equation applies, knowing the acquired speed we can compute the sufficient distance for the escape:
1 2mv 2 = GM m r r = 2GM v2 = 2 × 6.67 × 10 −11× 1.989 × 1030 (125×1036006)2 = 2.20 × 1011m
0.8
A satellite of mass 190 kg is placed into Earth orbit at a height of 700 km above the surface. • a) Assuming a circular orbit, how long does the satellite take to complete one orbit? • b) What is the satellite’s speed?.
• c) Starting from the satellite on the Earth’s surface, what is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet’s daily rotation.
a)
In problem 3), using Kepler’s Law for elliptic motion, we were able to get the semi-major axis distance in terms of the period of motion. Here we assume a simpler circular motion and we will use the radius instead (we should also
include the satellite height above the surface ): R3 = GM 4π2T 2 (700 × 103+ 6.37 × 106)3 = 6.67 × 10 −11× 5.972 × 1024 4π2 T 2 = 3.360 × 1018T2 T = 5918.15 s = 1.64 h
b)
Knowing the period and the radius we can easily derive the speed:
v = 2πR T = 2π(700 × 10 3+ 6.37 × 106) 5918.15 = 7507 m/s
c)
The total energy of the {Earth+Satellite} system is to be conserved:
• Initially, on earth surface and by taking earth’s speed as ve = 2π×6.37×106 24×3600 = 463.24 m/s Ei = 1 2mv 2 e− G M m r • On orbit, Ef = 1 2mv 2− GM m R
The change in energy is the minimum energy required to put this satellite on orbit: ∆E = 1 2m(v 2− v2 e) + GM m( 1 r− 1 R) = 1 2× 190 × (7507 2− 463.242) + 6.67 × 10−11× 5.972 × 1024× 190 × ( 1 6.37 × 106 − 1 6.37 × 106+ 700 × 103) = 6.52 × 109J
Studies of the relationship of the Sun to our galaxythe Milky Way have revealed that the Sun is located near the outer edge of the galactic disc, about 30 000 ly (1 ly = 9.46 ×1015
m) from the center. The Sun has an orbital speed of approximately 250 km/s around the galactic center.
• a) What is the period of the Sun’s galactic motion?
• b)What is the order of magnitude of the mass of the Milky Way galaxy?
• c) Suppose the galaxy is made mostly of stars of which the Sun is typical. What is the order of magnitude of the number of stars in the Milky Way?
a)
The period can be easily evaluated assuming a circular motion: v = 2πR T T = 2πR v = 2π × 30000 × 9.46 × 10 15 250 × 103 = 7.13267 × 1015s = 2.26 × 108yrs
b)
Using the same procedure as in problem 8-a): the period , the radius and the mass of the object giving rise to gravity are related with Kepler/Newton law:
R3 = GM 4π2T 2 M = 4π 2R3 GT2 = 4π 2× (30000 × 9.46 × 1015)3 6.67 × 10−11× (7.13267 × 1015)2 = 2.66 × 1041kg a 1041order of magnitude
c)
Such a huge mass will roughly include a number of Suns about 1030kg each:
an order of magnitude of 10104130 = 10
