# Significant Figures Verification

• TRB8985
TRB8985
Homework Statement
The gravitational pull of the Earth on an object is inversely proportional to the square of the distance of the object from the center of the Earth. At the Earth’s surface, this force is equal to the object’s normal weight, mg, where 𝑔 = 9.8 𝑚/𝑠^2 , and at large distances, the force is zero.

If a 20,000 kg asteroid falls to Earth from a very great distance away, what will be its minimum speed as it strikes the Earth’s surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the Earth’s atmosphere.
Relevant Equations
Kinetic energy imparted to planet = m_asteroid * g * radius_Earth
Good morning,

I've completed the problem provided above and have verified my answers are correct, but I'm running into a strange situation when it comes to the solution's answer for the kinetic energy portion.

For the kinetic energy imparted to the planet, we're taking 20,000 kg * 9.8 m/s² * 6.371E6 m. This completely matches the solution's answer.

I'm aware that there's one significant figure in 20,000, two in 9.8, and 4 in 6.371E6. My understanding is that when multiplying these values together, the result should be reported using the lowest number of significant figures in these three values - so just one, 1E12 J.
The official solution for this problem reports three, however - 1.25E12 J.

Am I making a mistake in my reporting of the answer? Or is the provided solution using the incorrect number of significant figures here?

Thank you!

Homework Helper
Homework Statement:: The gravitational pull of the Earth on an object is inversely proportional to the square of the distance of the object from the center of the Earth. At the Earth’s surface, this force is equal to the object’s normal weight, mg, where 𝑔 = 9.8 𝑚/𝑠^2 , and at large distances, the force is zero.

If a 20,000 kg asteroid falls to Earth from a very great distance away, what will be its minimum speed as it strikes the Earth’s surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the Earth’s atmosphere.
Relevant Equations:: Kinetic energy imparted to planet = m_asteroid * g * radius_Earth

Good morning,

I've completed the problem provided above and have verified my answers are correct, but I'm running into a strange situation when it comes to the solution's answer for the kinetic energy portion.

For the kinetic energy imparted to the planet, we're taking 20,000 kg * 9.8 m/s² * 6.371E6 m. This completely matches the solution's answer.
The energy for a fall from infinity to the surface under inverse square matches the energy for a fall from surface to center at constant surface gravity. Yes, that would be a correct result.

I'm aware that there's one significant figure in 20,000, two in 9.8, and 4 in 6.371E6. My understanding is that when multiplying these values together, the result should be reported using the lowest number of significant figures in these three values - so just one, 1E12 J.
I agree. One significant figure.
The official solution for this problem reports three, however - 1.25E12 J.
Unless the asteroid's mass has additional significant figures, I agree that this is too many significant figures.
Am I making a mistake in my reporting of the answer? Or is the provided solution using the incorrect number of significant figures here?
I agree with you and disagree with the book.

However, if one ascribes (questionably) three significant figures to that 20,000 kg then an argument can be made that three digits is appropriate in the result:

The maximum relative error in 9.8 (two sig figs) is only about one part in two hundred. Less than one part in one thousand in reality.

The relative error in 1.2 (two sig figs) is almost one part in twenty. About one part in 24 in this case.

Do we really want to be over-reporting the unreliability of the result by a factor of 4 (maximum error) or by a factor of 40 (actual error).

When presenting a result whose first digit is a 1 and whose least significant inputs have first digits of 8 or 9, I would tend to err on the side of adding one more significant digit in the result than the rules call for.

When adding large columns of numbers, I would tend to err on the side of subtracting one significant digit from the result. One more for every factor of 100 in the number of entries being totalled to account for statistical noise in the rounding error. But that's enough work that it may be worth doing a real error analysis at that point.

• TRB8985