1.
Determine, to two significant figures, the velocity of escape for earth of the celestials
−3 mile
bodies listed below. The data are rough and g may be taken to be 6.1 ×10 2
sec
Given:
Acceleration of Radius
Gravity at the (miles)
Surface
Venus 0.85 g 3 800
Mars 0.38 g 2 100
Jupiter 2.60 g 43 000
Sun 28 g 432 000
Ganymede 0.12 g 1 780
Solution:
For Venus,
V e =√ 2 gR
V e =√ 2(0.85)(6.1 ×10−3)(3 800)
∴ V e =6.30 miles/ sec answer
For Mars,
V e =√ 2 gR
V e =√ 2(0.38)(6.1 ×10−3)(2 100)
∴ V e =3.12 miles/sec answer
For Jupiter,
V e =√ 2 gR
V e =√ 2(2.60)(6.1 ×10−3)(43 000)
∴ V e =36.93 miles/sec answer
For Sun,
V e =√ 2 gR
V e =√ 2(28)(6.1 ×10−3)(432 000)
∴ V e =384.15 miles/sec answer
For Ganymede,
V e =√ 2 gR
V e =√ 2(0.12)(6.1 ×10−3)(1 780)
∴ V e =1.61 miles/sec answer
2. The radius of the moon is roughly 1080 miles. The acceleration of gravity at the
surface of the moon is about 0.165 g , where g is the acceleration of gravity at the
surface of the earth. Determine the velocity of escape for the moon.
Given:
gm=0.165 g
−3 mile
¿ 0.165(6.09× 10 2
)
sec
−3 mile
gm=1.00485 ×10 2
sec
Rm=1080 miles
Required: V e
Solution:
V e =√ 2(1.00485 ×10−3)(1080)
∴ V e =1.473 miles/sec answer
3. Determine the escape velocity of the Jupiter if its radius is 7149 Km and mass is
1.898 ×10²⁷ Kg.
Given:
Mass M =1.898 ×10²⁷ Kg
Radius R=7149 Km
Gravitational Constant G=6.67408 ×10−11m3 kg−¹ s−²
Solution:
Escape Velocity is givenas
Vesc=√ 2 GM / R
¿ √ 2 x 6.67408 ×10−¹¹ x 1.898×/7149
km
∴ 50.3 answer
s
4. Determine the escape velocity of the moon if Mass is 7.35 ×1022 Kg and the radius
is 1.5 ×10⁶ m.
Given:
M =7.35 ×1022 Kg ,
R=1.5× 106 m
Solution:
Escape Velocity formulais givenby
Vesc=√ 2 GMR
−11
¿ √ 2× 6.673 ×10 × 7.35× 1022/1.5× 10⁶
( )
5
10 m
∴ 7.59 answer
s
5. A thermometer reading 75 ℉ is taken out where the temperature is 20 ℉ . The
reading is 30 ℉ , 4 minutes later. Find (a) the thermometer reading 7 minutes after
thermometer was brought outside, and (b) the time taken for the reading to drop from
75 ℉ to within a half degree of the air temperature.
Given:
t u ua
0 75 ℉ 20 ℉
4 30 ℉ 20 ℉
7 ? 20 ℉
? 20.5 ℉ 20 ℉
Solution:
−kt
u=ua +c e
−kt
u=20+c e
@t=0 ,u=75 ℉
−k(0)
75=20+c e
∴ c=55
@t=4 , u=30℉
−k(4 )
30=20+55 e
4 30−20
(e ¿¿−k ) = ¿
55
−kt
u=20+55 c e
Therefore ,
( )
t
10 4
u=20+55
55
(a) @t=7 ,u=?
( )
t
10 4
u=20+55
55
∴ u=22.78 ℉ ≈ 23℉ answer
(b) @u=20.5 ℉ ,t=?
( )
t
10 4
20.5=20+55
55
( )
t
10 4 20.5−20
=
55 55
ln ( ) =ln
t
10 0.5
4
55 55
0.5
ln
t 55
=¿
4 10
ln
55
4 ln 0.5/55
t=
ln 10 /55
∴ t=11.03min ≈ 11.5min answer
6. A thermometer reading 18 ℉ is brought into a room where the temperature is 70 ℉ ;
1 minute later the temperature reading is 31 ℉ . Determine the temperature reading
as a function of time and in particular, Find the temperature reading 5 minutes after
the thermometer is first brought into the moon.
Given:
t u ua
0 18 ℉ 70 ℉
1 31 ℉ 70 ℉
Required:
a) uas a f ( t )
b) when t=5 , u?
Solution:
From General Equation:
−kt
u=ua +c e
−kt
u=70+c e
@t=0 ,u=18 ℉
−kt
18=70+c e
∴ c=−52
Therefore,
−kt
u=70+(−52)e
@t=1 , u=31 ℉
−k(1)
31=70−52 e
31−70
∴ e−k =
−52
−k 39
e =
52
Therefore,
( )
t
39
u=70+(−52)
52
or,
( )
t
−k 39
ln e =ln
52
∴ k=0.29
Substitute,
∴ u=70−52 e−( 0.29) t answer
@t=5 , u=?
− ( 0.29 ) (5)
u=70−52 e
∴ u=57.66 ℉ ≈ 58℉ answer
7. At 1:00 P.M. a thermometer reading 70 ℉ is taken outside where the air temperature
is −10 ℉ (10 below zero). At 1:02 P.M., the reading is 26 ℉ . At 1:05 P.M. , the
thermometer is taken back indoors where the air temperature is 70 ℉ . What is the
thermometer reading at 1:09 P.M.?
Given:
Outside:
t u ua
1:00 P.M. 0 70 ℉ −10 ℉
1:02 P.M. 2 26 ℉ −10 ℉
1:05 P.M. 5 ? −10 ℉
Inside
t u ua
1:05 P.M. 0 ? 70 ℉
1:09 P.M. 4 ? 70 ℉
Solution:
For outside:
−kt
u=ua +c e
−kt
u=−10+c e
@t=0 ,u=70 ℉
−kt
70=−10+c e
∴ c=80
And so,
−kt
u=−10+80 e
@t=2 , u=26 ℉
−k (2 )
26=−10+80 e
−k (2 ) 36
e =
80
( )
t
−k 36
e = 2
80
Therefore,
( )
t
36 2
u=−10+80
80
@t=5 , u=?
( )
5
36 2
u=−10+80
80
∴ u=0.8673 ℉ answer
For Inside:
−kt
u=ua +c e
−kt
u=70+c e
For the same temperature,
( )
t
−k 36
e = 2
80
Then,
( )
t
36 2
u=70+c
80
@t=4 , u=? (1:05 P.M.)
u=0.8673 ℉
( )
t
36 2
0.8673=70+
80
∴ c=−69.132
Therefore,
u=70+¿
@t=4 , u=? (1:09 P.M.)
( )
4
36
u=70−69.132 ¿ 2
80
∴ u=56.00 ℉ answer
8. At 9:00 A.M. a thermometer reading 70 ℉ is taken outdoors where the temperature
is 15 ℉ , At 9:05 A.M., the thermometer is 45 ℉ , At 9:10 A.M. the thermometer is
taken back indoors where the temperature is fixed at 70 ℉ . Find (a) the reading at
9:20 A.M.
Given:
Outdoors:
t u ua
9:00 A.M. 0 70 ℉ 15 ℉
9:05 A.M. 5 45 ℉ 15 ℉
9:10 A.M. 10 ? 15 ℉
t u ua
9:00 A.M. 0 u10 15 ℉
9:20 A.M. 10 ? 70 ℉
Solution:
Outdoors:
−kt
u=−10+c e
@t=0 ,u=70 ℉
−k (0)
70 ℉=−10+ c e
c=55
@t=5 , u=45 ℉
−k (5 )
45=15+55 e
( )
1
−k (5 ) 30
e = 5
55
Then,
( )
t
30 5
u=15+55
55
@t=10 , u=?
( )
10
30 5
u=15+55
55
∴ u10=31. 36℉
Indoors,
−kt
u=70+c e
For the same temperature,
( )
1
−k 30
e = 5
55
Then,
( )
1
30 5
u=70+c
55
@t=10 , u10=31. 36 ℉ (9:10 A.M.)
( )
0
30 5
31. 36=70+ c
55
∴ c=−38.63
Then,
( )
t
30 5
u=70−38.63
55
@t=10 , u=?(9:20 A.M.)
( )
10
30 5
u=70−38.63
55
∴ u=58.50 ℉ ≈ 58℉ answer
9. A thermometer reading 10 ℃ is brought in a room whose temperature is 18 ℃ . One
minute later, the thermometer reading is 14 ℃ . How long does it take until the
reading becomes 16 ℃ .
Given:
@t=0 ,T =10 ℃ , T m=18 ℃
@t=1 , T =14 ℃
@t=? , T =16 ℃
Solution:
kt
T =T m +c e
@t=0
k (0)
10=18+c e
c=−8
@t=1
k (1 )
14=18+(−8)e
k =−0.69315
@t=? , T =16 ℃
−0.69315 (t )
16=18−8 e
∴ t=2minutes answer
10. A metal is heated up to a temperature of 500 ℃ . It is then exposed to a temperature
of 38 ℃ . After 2 minutes, the temperature of the metal becomes 190 ℃. When will
the temperature of the metal be 100 ℃ ? What is the temperature of the metal after 4
minutes?
Given:
@t=0 ,T =500 ℃ , T m=38 ℃
@t=2 , T =190 ℃
@t=? , T =100℃
@t=4 , T =?
Solution:
kt
T =T m +c e
@t=0
k (0)
500=38+c e
c=462
@t=2
k(2)
190=38+462 e
k =−0.55581
@t=4
−0.55581(4 )
T =38+ 462 e
T =80.0009 ℃
@t=?
−0.55581 (t)
80.0009=38+462 e
∴ t=3.61332minutes answer
