Question

In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 43 m/s (about 96 mph, the speed of a good major league fastball)."
Required:
a. Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
b. Now calculate the maximum height using universal gravitation.
c. How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.

Answer 1

Answer:

a)$h_{max}=14536.16 m$

b)$h = 15687.9 m$

c)$PD=7.62\%$ The estimate is low.

Explanation:

a) Using the energy conservation we have:

$E_{initial}=E_{final}$

we have kinetic energy intially and gravitational potential energy at the maximum height.

$\frac{1}{2}mv^{2}=mgh_{max}$

$h_{max}=\frac{v^{2}}{2g}$

$h_{max}=\frac{43^{2}}{2*0.0636}$

$h_{max}=14536.16 m$  

b)  We can use the equation of the gravitational force

$F=G\frac{mM}{R^{2}}$   (1)

We have that:

$F = ma$    (2)

at the surface G will be:

$G=\frac{gR^{2}}{M}$

Now the equation of an object at a distance x from the surface.

is:

$F=\frac{mgR^{2}}{(R+x)^{2}}$

$m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}$

Using that dv/dt is vdx/dt and integrating in both sides we have:

$v_{0}=\sqrt{\frac{2gRh}{R+h}}$

$h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}$

$h=15687.9$

c) The difference is:

So the percent difference will be:

$PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%$

$PD=7.62\%$

The estimate is low.

I hope it helps you!