Book answer wrong? Work = Fd question. Space Shuttle entry

[Barclay]

Homework Statement

The space shuttle has 8.45 x 10(to the power12) joules of energy transfer over an 8000km flight path. What force is applied by the atmosphere?

Homework Equations

The book answer is 1.06 x 10 (to the power nine) Newtons.

I think the book answer is wrong. They've shown the calculation as 8.45 x 10(to the power12) / 8 x 10 (to the power 3).

I think the mistake is that they should have used the value 8 x 10 (to the power 3)

The Attempt at a Solution

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Work = Force x distance

Force = 8.45 x 10(to the power12) / 8000 x 10 (to the power 3).

Force = 1056250 N = 1.06 x 10 (to the power 6)

Please advise. This is High School physics

 

[DrClaude]

Yes, they appear to be off by 3 orders of magnitude.

By the way, you can use the x² symbol on the toolbar to get exponents: 10³.

 

[Barclay]

Yes, they appear to be off by 3 orders of magnitude.
By the way, you can use the x² symbol on the toolbar to get exponents: 10³.

Thanks ... so my answer Force = = 1.06 x 10⁶ is correct. I was thinking does the shuttle really have a flight path of 8000km as the question says? The distance from atmosphere to Earths surface is much less isn't it ?? Had a quick look on internet and looks like 400 km

 

[SteamKing]

Thanks ... so my answer Force = = 1.06 x 10⁶ is correct. I was thinking does the shuttle really have a flight path of 8000km as the question says? The distance from atmosphere to Earths surface is much less isn't it ?? Had a quick look on internet and looks like 400 km

Yes, but does the shuttle come to a complete stop in outer space and then fall straight to the ground? That's the only way the distance on re-entry becomes 400 km.

 

[Barclay]

Yes, but does the shuttle come to a complete stop in outer space and then fall straight to the ground? That's the only way the distance on re-entry becomes 400 km.

Oh I see ... the shuttle does not just drop onto Earth perpendicular to a point on the surface. It flies in an orbit around the Earth so the path is 8000 km

 

[SteamKing]

Oh I see ... the shuttle does not just drop onto Earth perpendicular to a point on the surface. It flies in an orbit around the Earth so the path is 8000 km

You got it, except the Shuttle is technically no longer in orbit.

The great length of this path is designed so that the Shuttle bleeds off the kinetic energy from orbiting the Earth and slows down to a manageable landing speed by the time it reaches its landing field. The Shuttle is not powered after re-entry to the atmosphere and glides to its final destination.