Question

Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9×1022kg and making the assumption that its mass per unit volume is the same as Earth's.

Answer 1

Answer:$g=1.97 m/s^2$

Explanation:

Given

mass of Jupiter is $M=4.9\times 10^{22} kg$

Density of Jupiter is same as Earth

$density\ of\ Earth=density\ of\ Jupiter=5510 kg/m^3$

$mass=volume\times density$

considering Jupiter to be sphere of radius r

$M=\frac{4}{3}\times \pi r^3\times \rho$

$r^3=\frac{3M}{\rho \times 4\pi}$

$r^3=\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi}$

$r=(\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi})^{\frac{1}{3}}$

$r=1.28\times 10^6 m$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

$g=\frac{6.67\times 10^{-11}\times 4.9\times 10^{22}}{(1.28\times 10^6)^2}$

$g=1.97 m/s^2$

Answer 2

g = 1.97 m/s² is the acceleration due to gravity at the surface of Europa

What is acceleration due to the gravity?

It is the define as the acceleration gained by an object due to gravitational force. Its SI unit is m/s².

According to the question,

Mass of Jupiter = 4.9×10²² kg

Density of Earth = Density of Jupiter = 5520 kg/m³

If the radius of Jupiter is r then,

$M= \frac{4}{3} * \pi r^{3} * p\\\\r^{3} = \frac{3M}{p*4\pi } \\\\r^{3} = \frac{3*4.9*10^{22} }{5510*4\pi }\\\\r = (\frac{3*4.9*10^{22} }{5510*4\pi })^{\frac{1}{3} } \\\\r = 1.28 *10^{6}$

According to the Universal law of gravitation,

$g =\frac{GM}{r^{2} } = \frac{6.67*10^{-11}*4.9*10^{22} }{(1.28*10^{6})^{2} } \\g = 1.97 m/s^{2}$

Therefore,

The acceleration due to gravity at the surface of Europa is 1.97 m/s²

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