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A planet has a mass equal to (1/10)th(1/10)^{\text{th}} of Earth's mass and a diameter equal to half of Earth's diameter. The acceleration due to gravity on this planet is:

A

9.8 m s−29.8 \text{ m s}^{-2}

B

4.9 m s−24.9 \text{ m s}^{-2}

C

3.92 m s−23.92 \text{ m s}^{-2}

D

19.6 m s−219.6 \text{ m s}^{-2}

Step-by-Step Solution

  1. Formula: The acceleration due to gravity on the surface of a spherical body is given by g=GMR2g = \frac{GM}{R^2}, where GG is the gravitational constant, MM is the mass, and RR is the radius .
  2. Given Data:
  • Mass of the planet, Mp=110MEM_p = \frac{1}{10} M_E.
  • Diameter of the planet is half that of Earth, which implies the radius Rp=12RER_p = \frac{1}{2} R_E.
  • Acceleration due to gravity on Earth, gE=9.8 m s−2g_E = 9.8 \text{ m s}^{-2}.
  1. Ratio Calculation: gp=GMpRp2=G(ME/10)(RE/2)2g_p = \frac{G M_p}{R_p^2} = \frac{G (M_E / 10)}{(R_E / 2)^2} gp=110×1(1/2)2×GMERE2g_p = \frac{1}{10} \times \frac{1}{(1/2)^2} \times \frac{G M_E}{R_E^2} gp=110×4×gE=0.4gEg_p = \frac{1}{10} \times 4 \times g_E = 0.4 g_E
  2. Result: gp=0.4×9.8 m s−2=3.92 m s−2g_p = 0.4 \times 9.8 \text{ m s}^{-2} = 3.92 \text{ m s}^{-2}
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