Escape Orbit is On

Classical Mechanics Level 4

At a certain height ( = 3630 km =3630 \text{ km} ) from the surface of the earth, a spacecraft Rosetta orbits the earth in circular orbit with v orb ( = 6326 ms − 1 v_{\text{orb}} (=6326 \text{ ms}^{-1} ).

Meanwhile, at the surface of the earth, the ground control(for some reason) want to bring Rosetta, escape from its orbit around the earth(somewhere beyond the earth).

Find the approximated minimum impulse in Ns that Rosetta needs to exert to escape from its orbit around the earth!

Details and Assumptions

  • Gravitational constant G = 6.67 × 10 − 11 Nm 2 kg − 2 G= 6.67\times {{10}^{-11}} \text{ Nm}^2 \text{kg}^{-2} .
  • Mass of the earth M = 6 × 10 24 kg M = 6\times {{10}^{24}}\text{ kg} .
  • Mass of Rosetta m = 5 × 10 3 kg m=5\times {{10}^{3}} \text{ kg} .
  • Radius of the earth R = 6370 km R = 6370\text{ km} .
  • Neglect the gravity from other celestial objects.
  • Consider the two objects as two point of masses.
  • Neglect the change of m when Rosetta exerting impulse.
2106 × 10 4 2106\times {{10}^{4}} 3041 × 10 4 3041\times {{10}^{4}} 1053 × 10 4 1053\times {{10}^{4}} 2614 × 10 4 2614\times {{10}^{4}} 1311 × 10 4 1311\times {{10}^{4}}

Muhammad Daud by Muhammad Daud , 21, Indonesia

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1 solution

Muhammad Daud
Jan 16, 2016

First, let's begin with

I = m × Δ v I=m\times \Delta v …(1) , Where I is the impulse that Rosetta needs to exert.

Now, find the escape velocity at Rosetta’s orbit

v e s c = ( 2 G M r ) 1 2 {{v}_{esc}}={{\left( \frac{2GM}{r} \right)}^{\frac{1}{2}}} , Where r = R + h

*Notice that v e s c {{v}_{esc}} orbit will be equals to v e s c {{v}_{esc}} at the surface of the earth if r = R or h = 0

v e s c = ( 2 × 6.67 × 10 − 11 × 6 × 10 24 6370 × 10 3 + 3630 × 10 3 ) 1 2 {{v}_{esc}}={{\left( \frac{2\times 6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{6370\times {{10}^{3}}+3630\times {{10}^{3}}} \right)}^{\frac{1}{2}}}

v e s c ≈ 8946.5 {{v}_{esc}}\approx 8946.5

*if you want to make it faster, I put the true v o r b v_{orb} in the data, just use the relation between v o r b v_{orb} and v e s c {{v}_{esc}} v e s c = 2 × v o r b {{v}_{esc}}=\sqrt{2}\times {{v}_{orb}} *which will give us the same value as the calculation above.

Since Rosetta has initial velocity, hence Δ v = v e s c − v o r b \Delta v={{v}_{esc}}-{{v}_{orb}} …(2)

Then, from equation (1) and equation (2), we get

I = m ( v e s c − v o r b ) I=m({{v}_{esc}}-{{v}_{orb}})

I ≈ 5 × 10 3 ( 8946.5 − 6326 ) I\approx 5\times {{10}^{3}}(8946.5-6326)

I ≈ I\approx 13102500 \boxed{13102500} Ns

The closest answer is 1311 × 10 4 1311\times {{10}^{4}} Ns

1 Helpful 0 Interesting 0 Brilliant 0 Confused

yes i directly used root 2*orbital velocity

Manas Wadhwa - 5 years, 4 months ago

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