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College Physics

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

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BuyFindarrow_forward

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=250km

Radius of Earth, rEarth=6370km

Mass of ship, m=1×105kg

Formula used:

Formula to calculate the kinetic energy of the ship is,

  KEShip=GMEarthm2r    ...... (I)

Formula to calculate the potential energy of the ship is,

  PEShip=gmrEq    ...... (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,

  r=rEarth+rEq

Substitute 6370kmfor rEarthand 250kmfor rEqin the above expression.

  r=(6370km)+(250km)=6620km

Convert kminto m.

  r=6620×103m

Consider the data.

Gravitation constant, G=6.67×1011Nm2/kg2

Mass of Earth, MEarth=5.98×1024kg

Substitute 6.67×1011Nm2/kg2for G, 5.98×1024kgfor MEarth, 1×105kgfor m, and 6620×103mfor rin equation (I).

  KEShip=(6.67×1011Nm2/kg2)×(5.98×1024kg)×(1×105kg)2×(6620×103m)=3.988×10191320×103J=3.012×1012J

Substitute 250kmfor rEq, 1×105kgfor m, and 9.8m/s2for gin equation (II).

  PEShip=(9.8m/s2)×(1×105kg)×(250km)=(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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