If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be $(g on earth = 9.8 m / s e c^{2})$

4.9 m / s e c^{2}

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19.6 m / s e c^{2}

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8.9 m / s e c^{2}

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29.4 m / s e c^{2}

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Solution

The correct option is B $19.6 m / s e c^{2}$
Formula used: $g = \frac{G M}{R^{2}}$
Given,
$M_{p} = \frac{M_{e}}{2}$ and $R_{p} = \frac{R_{e}}{2}$

$g = \frac{G M}{R^{2}}$

$∴ g \propto \frac{M}{R^{2}}$

According to given situation,

$∴ \frac{g_{p}}{g_{e}} = (\frac{M_{p}}{M_{e}}) {(\frac{R_{e}}{R_{p}})}^{2} = (\frac{1}{2}) \times (2)^{2} = 2$

$\Rightarrow g_{p} = 2 g_{e} = 2 \times 9.8 = 19.6 m / s^{2}$

Final Answer: (c)

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Chandrayaan 1

Standard VIII Physics

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If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be (g on earth=9.8 m/sec2)

The correct option is B 19.6 m/sec2Formula used: g=GMR2 Given, Mp=Me2 and Rp=Re2 g=GMR2 ∴g ∝MR2 According to given situation, ∴gpge=(MpMe)(ReRp)2=(12)×(2)2=2 ⇒gp=2ge=2×9.8=19.6 m/s2 Final Answer: (c)

If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be $(g on earth = 9.8 m / s e c^{2})$