Energetics of orbits and escape velocity (HL)
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Flip to reveal answersKinetic energy of a body in a circular orbit?
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Question
Kinetic energy of a body in a circular orbit?
Answer
$KE = +\dfrac{GMm}{2r}$ — always **positive**.
Question
Gravitational potential energy of an orbiting body?
Answer
$PE = -\dfrac{GMm}{r}$ — **negative** (zero at infinity).
Question
Total energy of a body in a circular orbit?
Answer
$E = -\dfrac{GMm}{2r}$ — the sum KE + PE, and **negative** because the body is bound.
Question
How do |PE| and KE compare in an orbit?
Answer
The potential energy is **twice** the size of the kinetic energy: $|PE| = 2\,KE$.
Question
Why is total orbital energy negative?
Answer
Because the body is **bound** in the gravity well — you'd have to add energy (up to zero) to free it.
Question
What happens to E as the orbit gets higher?
Answer
$E = -GMm/2r$ becomes **less negative** (rises toward zero) — a higher, more weakly bound state.
Question
Define escape velocity.
Answer
The minimum launch speed that lets an object coast to infinity, ending with **zero** total energy.
Question
Escape velocity formula?
Answer
$v_{esc} = \sqrt{\dfrac{2GM}{r}}$.
Question
Does escape velocity depend on the launched mass?
Answer
**No** — the mass m cancels in ½mv² = GMm/r, so v_esc is the same for any object.
Question
How is escape velocity derived?
Answer
Set launch kinetic energy = depth of the well: ½mv² = GMm/r, then solve for v.
Question
Escape velocity from Earth's surface?
Answer
About $1.1\times10^4$ m s⁻¹ ≈ **11 km s⁻¹** (M = 6.0×10²⁴ kg, r = 6.4×10⁶ m).
Question
v_esc vs circular-orbit speed — what's the difference?
Answer
Escape velocity $\sqrt{2GM/r}$ carries a **factor of 2** under the root; orbital speed $\sqrt{GM/r}$ does not.
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