57. Escape Velocity According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from the earth's surface is F = mgR² (x + R)² where x = x(t) is the object's distance above the sur- face at time t, R is the earth's radius, and 9 is the accel- eration due to gravity. Also, by Newton's Second Law, F=ma m(dv/dt) and so dv m dt mgR² (x + R)² (a) Suppose a rocket is fired vertically upward with an initial velocity vo. Let h be the maximum height above the surface reached by the object. Show that 2gRh R+h [Hint: By the Chain Rule, m (dv/dt)= = mv (dv/dx).]
57. Escape Velocity According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from the earth's surface is F = mgR² (x + R)² where x = x(t) is the object's distance above the sur- face at time t, R is the earth's radius, and 9 is the accel- eration due to gravity. Also, by Newton's Second Law, F=ma m(dv/dt) and so dv m dt mgR² (x + R)² (a) Suppose a rocket is fired vertically upward with an initial velocity vo. Let h be the maximum height above the surface reached by the object. Show that 2gRh R+h [Hint: By the Chain Rule, m (dv/dt)= = mv (dv/dx).]
57. Escape Velocity According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from the earth's surface is F = mgR² (x + R)² where x = x(t) is the object's distance above the sur- face at time t, R is the earth's radius, and 9 is the accel- eration due to gravity. Also, by Newton's Second Law, F=ma m(dv/dt) and so dv m dt mgR² (x + R)² (a) Suppose a rocket is fired vertically upward with an initial velocity vo. Let h be the maximum height above the surface reached by the object. Show that 2gRh R+h [Hint: By the Chain Rule, m (dv/dt)= = mv (dv/dx).]
Hello I am stuck on this problem and not sure where I am going wrong in trying to show the V_0 expression. This problem is part of Calculus II residing in the separable differential equations section. Any help is appreciated, thank you.
Transcribed Image Text:A(L)= M
2(土)
= Ur
24
Le
25€
424
1.750
3A=M
1
A = M MARPT
3
[13] F = mg R ²
2
FIND Vo
-mg R4
2/2/17)
2
FgRZ
912
2
dv
ke
JA TJM = CMRE (JA-JM)
√A = (evke (TA-√m) -√m
JA=CETA - CoSm-5m
Fsma = m (3/4) sc, m (dv) = _mgR²
VU)=V
=
(x+R)²
WHEN, X=R, V = VO INITIAL CONDITION
PPOSITION
для
=mr(税)
XTR
X+R 2
dx = Svdv
=v
XHR 2
GRX
ZR = = V₁TC
gR
2
2
[2 = (31-216) —
2
-V²+C
WHEN X=L, V=U& CONDITION
2
V z 2g R g R + v v
2
дети?
0 = 2gR-gπtvo
24+4
gR- Zy R = v
21+4
2
gr (h+R) - 2gp voz
инг
др
gR²-c=2v²
x+17
ZAPR = (z(GR-V)) = v
AT MAN HEIGHT 2912
X+R -
r² = gRh+gR² - ZgR²
h+R
2
ZAR ²- GR+V EVZ
XHR
Transcribed Image Text:57. Escape Velocity According to Newton's Law of Universal
Gravitation, the gravitational force on an object of mass m
that has been projected vertically upward from the earth's
surface is
F =
mgR²
(x + R)²
where x = x(t) is the object's distance above the sur-
face at time t, R is the earth's radius, and 9 is the accel-
eration due to gravity. Also, by Newton's Second Law,
F= ma = m (dv/dt) and so
dv
m
dt
mgR²
(x + R)²
(a) Suppose a rocket is fired vertically upward with an
initial velocity vo. Let h be the maximum height above
the surface reached by the object. Show that
2gRh
R+ h
[Hint: By the Chain Rule, m (dv/dt) = mv (dv/dx).]
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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![A(L)= M
2(土)
= Ur
24
Le
25€
424
1.750
3A=M
1
A = M MARPT
3
[13] F = mg R ²
2
FIND Vo
-mg R4
2/2/17)
2
FgRZ
912
2
dv
ke
JA TJM = CMRE (JA-JM)
√A = (evke (TA-√m) -√m
JA=CETA - CoSm-5m
Fsma = m (3/4) sc, m (dv) = _mgR²
VU)=V
=
(x+R)²
WHEN, X=R, V = VO INITIAL CONDITION
PPOSITION
для
=mr(税)
XTR
X+R 2
dx = Svdv
=v
XHR 2
GRX
ZR = = V₁TC
gR
2
2
[2 = (31-216) —
2
-V²+C
WHEN X=L, V=U& CONDITION
2
V z 2g R g R + v v
2
дети?
0 = 2gR-gπtvo
24+4
gR- Zy R = v
21+4
2
gr (h+R) - 2gp voz
инг
др
gR²-c=2v²
x+17
ZAPR = (z(GR-V)) = v
AT MAN HEIGHT 2912
X+R -
r² = gRh+gR² - ZgR²
h+R
2
ZAR ²- GR+V EVZ
XHR](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b313955-89cd-4b2a-a000-54fbe962adf3%2F3e19081a-03f4-4645-9398-690c056c5777%2Fdore3x_processed.jpeg&w=3840&q=75)
![57. Escape Velocity According to Newton's Law of Universal
Gravitation, the gravitational force on an object of mass m
that has been projected vertically upward from the earth's
surface is
F =
mgR²
(x + R)²
where x = x(t) is the object's distance above the sur-
face at time t, R is the earth's radius, and 9 is the accel-
eration due to gravity. Also, by Newton's Second Law,
F= ma = m (dv/dt) and so
dv
m
dt
mgR²
(x + R)²
(a) Suppose a rocket is fired vertically upward with an
initial velocity vo. Let h be the maximum height above
the surface reached by the object. Show that
2gRh
R+ h
[Hint: By the Chain Rule, m (dv/dt) = mv (dv/dx).]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b313955-89cd-4b2a-a000-54fbe962adf3%2F3e19081a-03f4-4645-9398-690c056c5777%2Fcqurt3_processed.png&w=3840&q=75)
