Question

The surface of the sun has a temperature of about 5800K and consists largely of hydrogen atoms. Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67×10−27kg.) The escape speed for a particle to leave the gravitational influence of the sun is given by (2GM/R)1/2, where M is the sun's mass, R its radius, and G the gravitational constant. The sun`s mass is M=1.99×1030kg, its radius R=6.96×108m and G=6.673×10−11N⋅m2/kg2. Calculate the escape speed for the sun.

Answer 1

This question involves the concepts of kinetic energy, escape velocity, and rms speed.

a) The rms speed of hydrogen atom at 5800 K is "11991 m/s".

b) The escape velocity for the Sun is "6.18 x 10⁵ m/s".

a)

We will use the formula of the average kinetic energy of gas molecules to find out the rms speed of the hydrogen atom. rms speed is the root mean square speed of an atom:

$K.E = \frac{3}{2}KT=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{3KT}{m}}$

where,

v =rms speed = ?

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

T = absolute temperature = 5800 k

m = mass of hydrogen atom = 1.67 x 10⁻²⁷ kg

Therefore,

$v=\sqrt{\frac{3(1.38\ x\ 10^{-23}\ J/k)(5800\ k)}{1.67\ x\ 10^{-27}\ kg}}$

v = 11991 m/s

b)

The escape velocity of the Sun is given by the following formula:

$v_e=\sqrt{\frac{2GM}{R}}$

where,

ve = escape velocity = ?

G = Gravitational Constant = 6.673 x 10⁻¹¹ N.m²/kg²

M = Mass of Sun = 1.99 x 10³⁰ kg

R = radius of Sun = 6.96 x 10⁸ m

Therefore,

$v_e=\sqrt{\frac{2(6.673\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{6.96\ x\ 10^8\ m}}$

$v_e = 6.18\ x\ 10^5\ m/s$

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