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2026 USAAAO Third Exam (Theory)1 Instructions (Please Read Carefully)Условие Решение

USAAAO 2026, III кръг — Задача 1

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2026 USAAAO Third Exam (Theory) · 10 т.

Instructions (Please Read Carefully)

2026 USAAAO Third Exam (Theory)

1 Instructions (Please Read Carefully)

The test must be completed within 3.5 hours (210 minutes) and the maximum number of points is 260 points. The point breakdown per problem is as follows.

Problem	P1	P2	P3	P4	P5	Total
Points	10	20	50	60	120	260

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. Some problems may require a designated answer sheet; these will be provided. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific calculator (non-programmable and non-graphing) for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on June 25, 2026. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Students must be proctored during the entire duration of the exam.

This exam sheet and your answer sheets (including scratch papers) should be returned to your proctors once the exam ended. Your proctors should upload your answer sheets to the Google Form provided to them.

We acknowledge the following people for their contributions to this year’s theory exam:

Feodor Yevtushenko, Hagan Hensley, Joe McCarty, Lucas Pinheiro

After reading the instructions, please make sure to sign, affirming that:

All work on this exam has been completed by me.
I took this exam under the supervision of a proctor.
I did not receive any external help beyond the materials provided.
I will not discuss the contents of this exam with anyone until June 24, 2026.
Failure to follow these rules will result in disqualification from the exam.

Student Signature: ____________________ Date: ____________________

Условие

For both parts of this question, assume Earth’s current orbit is circular with a radius $a=1\ \mathrm{AU}$ .

(a) (3 points) If the Sun were to instantly lose 40% of its mass, what would happen to Earth’s orbit? Assume that the mass simply disappears without interacting with the Earth. Find the semi-major axis $a$ and the eccentricity $e$ of the new orbit.

(b) (7 points) Instead consider the scenario where this same 40% mass loss happens gradually over millions of years, as it will when the Sun is near the end of its life. In this scenario, find the semi-major axis $a$ and the eccentricity $e$ of the new orbit.

Решение

Solution:

(a) The new mass of the Sun is $0.6M_\odot$ . The Earth is still moving with its original orbital velocity $v=\sqrt{GM_\odot/(1\ \mathrm{AU})}$ immediately after the Sun loses its mass, but this is now faster than the velocity required for a circular orbit. Vis-viva gives

v^2=\frac{GM_\odot}{1\ \mathrm{AU}}=0.6GM_\odot\left(\frac2{1\ \mathrm{AU}}-\frac1a\right)

a=\frac1{\dfrac2{1\ \mathrm{AU}}-\dfrac1{0.6\times1\ \mathrm{AU}}}=\boxed{3\ \mathrm{AU}}\quad\text{(1.5 points)}

The perihelion distance should still be 1 AU, so we also have $e=1-\frac{1\ \mathrm{AU}}{3\ \mathrm{AU}}=\boxed{0.67}$ (1.5 points).

(b) This process occurs much more slowly than Earth’s orbital period, so there’s nothing to break the symmetry of the initially circular orbit as in part (a). Thus, the orbit should remain circular and gradually move outwards as the Sun loses its mass. Since the orbit is circular we have $\boxed{e=0}$ (1 point).

Conservation of energy does not apply here, since this is an open system losing mass to its environment. However, conservation of the Earth’s angular momentum should still hold (2 points for realizing this). We can justify this more carefully using the (classical) adiabatic theorem, which in this case says that the Earth’s angular momentum is an adiabatic invariant: it is unchanged for very slow changes in the system Hamiltonian.

If $M_\oplus\ll M_\odot$ and the Earth has a circular orbit with semimajor axis $a$ , then the Earth’s angular momentum is

L=M_\oplus av=M_\oplus\sqrt{GM_\odot a}\quad\text{(1 point)}

This quantity must be conserved. Since $M_\oplus$ and $G$ don’t change, this implies $M_\odot a$ must be constant as $M_\odot$ drops (1 point). Thus we have

\frac a{1\ \mathrm{AU}}=\frac{M_\odot}{0.6M_\odot}

\boxed{a=1.67\ \mathrm{AU}}\quad\text{(2 points)}

This process will happen in several billion years when the Sun expands into a red giant and loses much of its mass to solar wind. The Earth’s expanded orbit may save it from being swallowed by the red giant Sun, though the exact outcome remains an open question.

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