Which launch velocity is greater?

Classical Mechanics Level 2

Simplifying assumptions:

  • Vacuum conditions.
  • Launch from surface of spherical, homogeneous density object, S .
  • The test particles are immediately at launch velocity vertical upwards normal to surface of object S .

Case 1: (quite non physical): the gravitational field is at S surface value to 1 S radius. What is the necessary launch velocity (V1) needed to reach just that altitude with 0 velocity upon reaching there.

Case 2: the gravitational field follows normal physics and declines on an inverse square basis measured in S radii. What is the necessary launch velocity (V2) needed so that the test particle's velocity is asymptotic to 0 (that is, never 0, but its limit is 0).

V1 = V2 V1 < V2 The answer has to be something else! V1 > V2

A Former Brilliant Member by A Former Brilliant Member

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1 solution

∫ 1 r 2 d r ⟹ − 1 r \int \frac{1}{r^2} \, dr \Longrightarrow -\frac{1}{r} . ∫ 1 ∞ 1 r 2 d r ⟹ 1 \int_1^{\infty } \frac{1}{r^2} \, dr \Longrightarrow 1 . The velocities are equal.

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