Chapter 28, Problem 68PE

### College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000

Chapter
Section

### College Physics

1st Edition
Paul Peter Urone + 1 other
ISBN: 9781938168000
Textbook Problem

# Nuclear-powered rockets were researched for some years before safety concerns became paramount. (a) What fraction of a rocket's mass would have to be destroyed to get it into a low Earth orbit, neglecting the decrease in gravity? (Assume an orbital altitude of 250 km, and calculate both the kinetic energy (classical) and the gravitational potential energy needed.) (b) If the ship has a mass of 1 .00×10 5 kg (100 tons), what total yield nuclear explosion in tons of TNT is needed?

To determine

(a)

The fraction of mass of rocket's fuel that need to destroy to get the rocket into low Earth orbit when neglecting the decrease in gravity.

Explanation

Given:

Distance of ship from the equator, rEq=250â€‰km

Mass of ship, m=1Ã—105â€‰kg

Formula used:

Formula to calculate the kinetic energy of the ship is,

â€ƒâ€ƒKEShip=GMEarthm2râ€ƒâ€ƒâ€ƒâ€ƒ...... (I)

Formula to calculate the potential energy of the ship is,

â€ƒâ€ƒPEShip=gmrEqâ€ƒâ€ƒâ€ƒâ€ƒ...... (II)

Calculation:

Since, the distance of the ship from the center of Earth is the sum of the radius of Earth and the distance of the ship from equator.

The distance of the ship from the center of Earth is,

â€ƒâ€ƒr=rEarth+rEq

Substitute 6370â€‰kmfor rEarthand 250â€‰kmfor rEqin the above expression.

â€ƒâ€ƒr=(6370â€‰km)+(250â€‰km)=6620â€‰km

Convert kminto m.

â€ƒâ€ƒr=6620Ã—103â€‰m

Consider the data.

Gravitation constant, G=6.67Ã—10âˆ’11â€‰Nâ‹…m2/kg2

Mass of Earth, MEarth=5.98Ã—1024â€‰kg

Substitute 6.67Ã—10âˆ’11â€‰Nâ‹…m2/kg2for G, 5.98Ã—1024â€‰kgfor MEarth, 1Ã—105â€‰kgfor m, and 6620Ã—103â€‰mfor rin equation (I).

â€ƒâ€ƒKEShip=(6.67Ã—10âˆ’11â€‰Nâ‹…m2/kg2)Ã—(5.98Ã—1024â€‰kg)Ã—(1Ã—105â€‰kg)2Ã—(6620Ã—103â€‰m)=3.988Ã—10191320Ã—103â€‰J=3.012Ã—1012â€‰J

Substitute 250â€‰kmfor rEq, 1Ã—105â€‰kgfor m, and 9.8â€‰m/s2for gin equation (II).

â€ƒâ€ƒPEShip=(9.8â€‰m/s2)Ã—(1Ã—105â€‰kg)Ã—(250â€‰km)=(9

To determine

(b)

The total yield of nuclear explosion of TNT to get the rocket into a lower Earth orbit.

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