Answer:
a)[tex]h_{max}=14536.16 m[/tex]
b)[tex]h = 15687.9 m[/tex]
c)[tex]PD=7.62\%[/tex] The estimate is low.
Explanation:
a) Using the energy conservation we have:
[tex]E_{initial}=E_{final}[/tex]
we have kinetic energy intially and gravitational potential energy at the maximum height.
[tex]\frac{1}{2}mv^{2}=mgh_{max}[/tex]
[tex]h_{max}=\frac{v^{2}}{2g}[/tex]
[tex]h_{max}=\frac{43^{2}}{2*0.0636}[/tex]
[tex]h_{max}=14536.16 m[/tex]
b) We can use the equation of the gravitational force
[tex]F=G\frac{mM}{R^{2}}[/tex] (1)
We have that:
[tex] F = ma [/tex] (2)
at the surface G will be:
[tex]G=\frac{gR^{2}}{M}[/tex]
Now the equation of an object at a distance x from the surface.
is:
[tex]F=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
[tex]m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
Using that dv/dt is vdx/dt and integrating in both sides we have:
[tex]v_{0}=\sqrt{\frac{2gRh}{R+h}}[/tex]
[tex]h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}[/tex]
[tex]h=15687.9[/tex]
c) The difference is:
So the percent difference will be:
[tex]PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%[/tex]
[tex]PD=7.62\%[/tex]
The estimate is low.
I hope it helps you!
