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Difficulty: MediumGravitational Field and Orbits

An object of mass mm is projected vertically upwards from the surface of a spherical planet of radius RR and mass MM with an initial speed equal to half of the planet's escape velocity. Neglecting atmospheric friction, what maximum height above the planet's surface will the object reach?

  1. A
    R4\frac{R}{4}
  2. R3\frac{R}{3}Answer
  3. C
    R2\frac{R}{2}
  4. D
    RR

Answer

The maximum height above the planet's surface reached by the object is R3\frac{R}{3}.
By mechanical energy conservation, the initial total energy at launch equals the potential energy at maximum height where velocity is zero. The launch velocity v=12ve=122GMRv = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}} gives an initial kinetic energy of GMm4R\frac{GMm}{4R}. Combined with initial potential energy −GMmR-\frac{GMm}{R}, total energy is −3GMm4R-\frac{3GMm}{4R}. Equating this to −GMmR+h-\frac{GMm}{R+h} gives R+h=43RR+h = \frac{4}{3}R, yielding a height above the surface of h=R3h = \frac{R}{3}.

Step-by-Step Solution

1
Express the initial speed in terms of gravitational constant GG, mass MM, and radius RR.
The escape velocity is ve=2GMRv_e = \sqrt{\frac{2GM}{R}}. Thus, launch speed v=12ve=122GMRv = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}.
Escape velocity is defined as the minimum speed needed to escape the gravitational field.
2
Calculate the initial total mechanical energy at the planet's surface.
Initial kinetic energy Ki=12mv2=12m(2GM4R)=GMm4RK_i = \frac{1}{2}m v^2 = \frac{1}{2}m\left(\frac{2GM}{4R}\right) = \frac{GMm}{4R}. Surface potential energy Ui=−GMmRU_i = -\frac{GMm}{R}. Total energy Ei=Ki+Ui=GMm4R−GMmR=−3GMm4RE_i = K_i + U_i = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}.
Total mechanical energy is the sum of kinetic energy and gravitational potential energy.
3
Apply energy conservation to find the maximum distance from the planet's center.
At maximum height hh, speed is zero (Kf=0K_f = 0), so distance from center is r=R+hr = R + h. Energy Ef=−GMmR+hE_f = -\frac{GMm}{R+h}. Equating Ei=EfE_i = E_f gives −3GMm4R=−GMmR+h  ⟹  R+h=43R  ⟹  h=R3-\frac{3GMm}{4R} = -\frac{GMm}{R+h} \implies R+h = \frac{4}{3}R \implies h = \frac{R}{3}.
Mechanical energy is conserved in a central gravitational force field.

Key Concept

Conservation of Mechanical Energy in a Gravitational Field
Estimated Time:1m 30s
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Topics

Gravitational Field and Orbits

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