Question

y(t) = yg +ve+-0.5gt Meteor! y = 7,000 km 1) Show that a meteor, initially at rest (V,0 = 0) will take about 20 minutes to fall from an altitude of 7,000 km as shown in the diagram to the right. Note that the change in y is negative and you will need to convert kom into meters for the units to work out correctly. y = 0 km R= 6,371 km The calculation from problem 1) is incorrect though. The Earth reason being that acceleration is not constant. The acceleration due to gravity varies with height. When near the surface of the Earth objects accelerate at g = 9.8 m/s? however when the meteor is further away from the Earth it's acceleration is less. The meteors acceleration is actually given by the formula. M Where G=-6.67x10-1N. (The gravitational constant) and kg M=5.97x1024 kg (the mass of the Earth) and r= the distance from the center of the Earth to the objects location. 2) Create a graph using Mathematica that plots the acceleration due to gravity between y = 0 km and y = 7,000 kom. 3) Is your graph consistent with the gravity at Earth's surface being approximately g=9.8 m/s (it should be) dv, 4) Knowing that a, = use Euler's Method to calculate the velocity of the meteor as a function of time between y = 0 dt km and y = 7,000 km. (assume the velocity is initially zero) Use At =1.0 sec = dy 5) Knowing , use Euler's Method to calculate the height of the meteor as function of time between y = 0 km and dt y = 7,000 kom. Use At=1.0 sec 6) Create a graph of your results from both 4 and 5. 7) Does your calculation from part 5 indicate the meteor will take more time or less time to fall then your result from 1)?

Answer 1

Solution:

(1)

Given,

Hieght of the meteor $h=7000km=7000\times 10^3m$

Meteor is initially at rest, $u=0$ .

Take acceleration due to gravity $g=9.8m/s^2$

Let the time taken to fall from the height be t.

By using equation of motion,

$h=ut+\frac{1}{2}gt^2$

Put, $u=0$

  

h = =gt

So,

$t=\sqrt{\frac{2\times h}{g}}=\sqrt{\frac{2\times 7000\times 10^3}{9.8}}$

$t=1195.23 sec=19.92min\approx 20min$

Thus time taken by the meteor is 20 min.