PHYSICS 111 HOMEWORK SOLUTION #13

PHYSICS 111 HOMEWORK SOLUTION #13

May 1, 2013

0.1

In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational 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:

mM F = G r2

=

6.67

?

10-11

?

2.10 ? 21 ? 10-3 (3.90 ? 10-2)2

= 1.93 ? 10-3 N

2

0.2.

0.2

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

=

G

mM r2

and

the

free-fall force mg

:

mM mg = G r2

M g = G r2

=

6.67

?

10-11

?

6.68 ? 1019 (242 ? 103)2

= 0.0761 m/s2

3

b)

We can use the free-fall equation of motion under the above-calculated accel-

eration

h

=

1 2

gt2

to

get

the

time

in

flight:

2h t=

g

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:

vy2 - vy20 = 2gh

vy

=

2gh

= 2 ? 0.0761 ? 5000

= 27.59 m/s

and

v = 27.592 + 7.702 = 28.6 m/s

The direction of his impact is :

= arctan vy vx 27.59

= arctan 7.7

= 74.4

4

0.3.

0.3

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 42

:

a3

=

GM 42

T

2

=

6.76

?

10-11 ? 1.989 42

?

1030

T

2

= 3.360 ? 1018T 2

= 3.360 ? 1018 ? (90.6 ? 365 ? 24 ? 3600)2

= 2.75833 ? 1037 m3

5

and

a = 3.02145 ? 1012 m 3.02145 ? 1012

= 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

G

Mn m r2

which

is

balanced

by

the

centripetal

mv2 r

force

that

keeps

the

star

in its orbit:

G

Mnm r2

=

mv2 r

m(r)2 =

r

=

GMn r3

=

6.76

?

10-11

?

2

?

1.989 ? 110003

1030

= 14122 rad/s

6

0.5.

0.5

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 GM m

= -(-

-

)

R

GM m

= -(-

- 0)

R

6.67 ? 10-11 ? 7.36 ? 1022 ? 995

=

1.74 ? 106

= 2.81 ? 109 J

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? 1030kg but will have a smaller radius comparable to earth's r = 6.37 ? 106m.

7

The density is thus:

=

Ms

4r3

3

1.989 ? 1030 = 4?(6.37?106)3

3

= 1.84 ? 109 kg/m3

b)

Following the same steps in problem 2-a), the free fall acceleration is :

M g = G r2

=

6.67

?

10-11

?

1.989 ? 1030 (6.37 ? 106)2

= 3.27 ? 106 m/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 ? 1030 ? 6.37 ? 106

4.93

= -1.03 ? 1014 J

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 r

m

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 2

mv2

comparable

to

the

first

amount:

8

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