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Difficulty: MediumGravitational Field and Orbits

A uniform spherical planet has twice the mass and twice the radius of the Earth. If the acceleration due to gravity at the Earth's surface is gg, what is the acceleration due to gravity at the surface of this planet?

  1. A
    0.25g0.25g
  2. 0.5g0.5gAnswer
  3. C
    1.0g1.0g
  4. D
    2.0g2.0g

Answer

The acceleration due to gravity at the surface of the planet is 0.5g0.5g.
The acceleration due to gravity at a planet's surface is given by g=GMR2g = \frac{GM}{R^2}. Doubling the mass doubles the field strength, but doubling the radius reduces the field strength by a factor of 22=42^2 = 4 due to the inverse-square law. Combining these changes results in 24g=0.5g\frac{2}{4}g = 0.5g.

Step-by-Step Solution

1
Write the general expression for surface acceleration due to gravity.
g=GMR2g = \frac{GM}{R^2}, where GG is the gravitational constant, MM is planetary mass, and RR is planetary radius.
Establishes the fundamental formula relating gravity to mass and radius.
2
Substitute the given parameters for the new planet (Mp=2MM_p = 2M and Rp=2RR_p = 2R).
gp=G(2M)(2R)2=2GM4R2g_p = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2}.
Applies the proportional changes to mass and radius into the formula.
3
Simplify the expression to find gpg_p in terms of gg.
gp=24(GMR2)=0.5gg_p = \frac{2}{4} \left(\frac{GM}{R^2}\right) = 0.5g.
Evaluates the fractional change relative to Earth's surface gravity.

Key Concept

Dependence of surface gravitational field strength on planetary mass and radius via the inverse-square law.
Estimated Time:1m 15s
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Topics

Gravitational Field and Orbits

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