What monatomic ions do barium (Z 56) and selenium (Z 34)

Chapter 2, Problem 2.71

(choose chapter or problem)

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

                                               \(F=\frac{m g R^{2}}{(x+R)^{2}}\)

Where \(x=x(t)\) is the object's distance above the surface at time \(t,R\) is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F=m a=m(d v / d t)\) and so

                                                \(m \frac{d y}{d t}=\frac{m g R^{2}}{(x+R)^{2}}\)

(a) Suppose a rocket is fired vertically upward with an initial velocity \(v_{0}\) Let \(h\)  be the maximum height above the surface reached by the object. Show that

                                                  \(v_{0}=\sqrt{\frac{2 g R h}{R+h}}\)

[Hint: By the Chain Rule, \(m(d v / d t)=m(d v / d t)\).]

(b) Calculate \(v_{e}=\lim _{h \rightarrow \infty} v_{0}\). This limit is called the escape velocity for the earth. (Another method of finding escape velocity is given in Exercise 7.8.77.)

(c) Use \(R=3960 \mathrm{mi}\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) to calculate \(v_{e}\) in feet per second and in miles per second.

Equation Transcription:

 

 

 

 

Text Transcription:

m

F= mgR^2/(x+R)^2

x=x(t)

g

t,R

F=m a=m(d v / d t)

mdy/dt=mgR^2/(x+R)^2

v sub 0

h

v sub 0 = square root 2gRh/R+h

m(dv/dt) = m (dv/dt)

v sub e = lim_h right arrow infinity v sub 0

R= 3960 mi

g= 32ft/s^2

v sub e