Classical mechanicsLesson 3 of 7 in the course Gravitation

Gravitational Energy, Escape Speed, and Bound Motion

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

Near Earth’s surface, we can often treat gravitational acceleration gg as constant and use ΔU=mgΔy\Delta U = mg\Delta y. Across large distances, however, gravity weakens with distance, so this approximation is not sufficient. This lesson develops the gravitational potential energy formula that works beyond Earth’s surface.

The formula ΔU=mg(y2−y1)\Delta U = mg(y_2-y_1) assumes that gravitational acceleration stays approximately constant between the two heights. That is a good approximation for modest changes in height near Earth’s surface. It becomes less accurate when an object travels hundreds or thousands of kilometers, because the gravitational force decreases as the distance from Earth’s center increases.

All flashcards

What is gravitational potential energy when it is zero at infinite separation?
U=−GMm/rU=-GMm/r, where rr is the separation between masses MM and mm.
How is gravitational potential energy change calculated between two separations?
ΔU=GMm(1/r1−1/r2)\Delta U=GMm(1/r_1-1/r_2).
What does negative total mechanical energy indicate in the ideal gravitational model?
The object is gravitationally bound and cannot reach infinity.
What is escape speed from distance rr from a spherical body?
vesc=2GM/rv_{esc}=\sqrt{2GM/r}; it is independent of the escaping object's mass.
What characterizes a conservative force?
Its work between two endpoints is independent of the path taken.

Quiz questions

  1. What is the magnitude of the gravitational force between masses MM and mm separated by distance rr?Answer: F=GMm/r2F=GMm/r^2
  2. Why does ΔU=mgΔy\Delta U=mg\Delta y become inaccurate over large changes in distance from a planet?Answer: Gravitational acceleration changes significantly as distance from the planet changes.
  3. Why does gravity do no work on an object moving along a circular arc centered on a planet?Answer: Gravity points radially, perpendicular to the displacement along the arc.
  4. Under ideal conditions, what does surface escape speed represent?Answer: The minimum launch speed for reaching infinity with zero speed.
  5. Why does the gravitational potential energy increase for a payload raised from a planet's surface not give the full energy needed to place it into orbit?Answer: It excludes orbital kinetic energy, launcher energy, and practical losses.

Based on University Physics Volume 1 by William Moebs, Samuel J. Ling, Jeff Sanny (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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