## Question

Imagine a planet whose diameter and mass are both one-half of those of earth. The day’s surface temperature of this planet reaches upto 800 K. Are oxygen molecules possible in the atmosphere of this planet? Give calculation. (Escape velocity on earth’s surface = 11.2 km s^{–1}, Boltzmann’s constant

*k* = 1.38 × 10^{–23} JK^{–1}, mass of oxygen molecule = 5.3 × 10^{–26 }kg.)

### Solution

0.79 km s^{–1}

The escape velocity of a body from earth’s surface is

Where *M _{e}* and

*R*are mass and radius of the earth respectively. If mass of the given planet be

_{e}*M*and radius

_{P}*R*, then the escape velocity from the planet is given by

_{P}

The translation kinetic energy of an oxygen molecule at temperature *T* is Therefore, if *m* be the mass and *v* the thermal velocity of an oxygen molecule, we have

This velocity is less than the escape velocity. Hence oxygen molecules are possible in the atmosphere of this planet.

#### SIMILAR QUESTIONS

If the period of revolution of an artificial satellite just above the earth be *T*and the density of earth be then prove that ρT^{2} is a universal constant. Also calculate the value of this constant.

A space-craft is launched in a circular orbit near the earth. How much more velocity will be given to the space-craft so that it will go beyond the attraction force of the earth. (Radius of the earth = 6400 km, *g* = 9.8 m/s^{2}).

An artificial satellite of mass 200 kg revolves around the earth in an orbit of average radius 6670 km. Calculate its orbital kinetic energy, the gravitational potential energy and the total energy in the orbital.

(Mass of earth = 6.0 × 10^{24} kg, *G* = 6.67 × 10^{–11} Nm^{2} kg^{ –2}).

With what velocity must a body be thrown upward from the surface of the earth so that it reaches a height of 10 *R _{e}*? Earth’s mass and

*G*= 6.67 × 10

^{ –11}Nm

^{2}kg

^{–2}.

A rocket is launched vertically from the surface of the earth with an initial velocity of 10 km s^{–1}. How far above the surface of the earth would it go? Mass of the earth = 6.0 ×^{ }10^{24} kg, radius = 6400 km and *G* = 6.67 ×^{ }10^{ –11} Nm^{2} kg^{ –2}. ^{ }

The escape velocity of a body from earth is 11.2 km s^{–1}. If the radius of a planet be half the radius of the earth and its mass be one-fourth that of earth, then what will be the escape velocity from the planet?

A body is at a height equal to the radius of the earth from the surface of the earth. With what velocity be it thrown so that it goes out of the gravitational field of the earth? Given:

N m^{2} kg^{–2}.

A particle falls on the surface of the earth from infinity. If the initial velocity of the particle is zero and friction due to air is negligible, find the velocity of the particle when it reaches the surface of the earth. Also find its kinetic energy. (Radius of earth is 6400 km and *g* is 9.8 m/s^{2}.)

A mass of is to be compressed in the form of a sphere. The escape velocity from its surface is What should be the radius of the sphere. Gravitational constant

A hollow spher is made of a lead of radius *R* such that its surface touches the outside surface of the lead sphere and passes through its centre. The mass of the lead sphere before hollowing was *M*. What is the force of attraction that this sphere would exert on a particle of mass *m *which lies at a distance from the centre of the lead sphere on the straight line joining the centres of the sphere and the hollow (as shown in fig.)?