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:

F

2

mM

r2

=

G

=

6.67 ¡Á 10?11 ¡Á

=

1.93 ¡Á 10?3 N

2.10 ¡Á 21 ¡Á 10?3

(3.90 ¡Á 10?2 )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 ¡Á1019 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

r 2 and the free-fall force mg :

mg

g

mM

r2

M

= G 2

r

= G

=

6.67 ¡Á 10?11 ¡Á

=

0.0761 m/s2

6.68 ¡Á 1019

(242 ¡Á 103 )2

3

b)

We can use the free-fall equation of motion under the above-calculated acceleration ?h = 12 gt2 to get the time in flight:

s

t =

2?h

g

r

2 ¡Á 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 = vx t:

x

=

7.70 ¡Á 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:

2

vy2 ? vy0

=

vy

=

=

2g?h

p

2g?h

¡Ì

2 ¡Á 0.0761 ¡Á 5000

=

27.59 m/s

and

v

=

p

=

28.6 m/s

27.592 + 7.702

The direction of his impact is :

¦È

4

vy

vx

27.59

= arctan

7.7

= 74.4?

=

arctan

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 ¡Á1011 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

4¦Ð 2 :

a3

GM 2

T

4¦Ð 2

6.76 ¡Á 10?11 ¡Á 1.989 ¡Á 1030 2

=

T

4¦Ð 2

= 3.360 ¡Á 1018 T 2

=

=

3.360 ¡Á 1018 ¡Á (90.6 ¡Á 365 ¡Á 24 ¡Á 3600)2

=

2.75833 ¡Á 1037 m3

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download