Escape Speed

Escape Velocity and Density Relation

The escape velocity from a spherical body is given by the formula:

Where:

• is the gravitational constant,
• is the mass of the body,
• is the radius of the body.

Substituting the expression for mass in terms of density (where ), we get:

This simplifies to:

Conclusion:

Therefore, the ratio of escape velocities for two bodies with different radii and densities is:

To solve the problem of finding the escape velocity from the surface of another planet with characteristics compared to Earth, we need to understand the formula for escape velocity and how changes in planetary characteristics affect it.

The escape velocity from the surface of a planet is given by the formula:

Where:

• is the escape velocity,
• is the universal gravitational constant,
• is the mass of the planet,
• is the radius of the planet.

For the Earth, the escape velocity is given as . So,

For the other planet, it is mentioned that its radius is four times that of Earth and it has the same mass density. Let be the mass density:

and .

If the radius of the new planet is , its volume is:

Therefore, the mass is:

Now, substitute and back into the escape velocity equation to find the escape velocity for the new planet:

Simplifying, we get:

Thus, the escape velocity from the surface of the new planet is . The correct answer is 4v.

Minimum Energy Required to Launch a Satellite

The minimum energy required to launch a satellite into orbit is equal to the work done in moving the satellite from the Earth's surface to its orbit. This energy is calculated based on the change in gravitational potential energy and kinetic energy.

Step 1: Gravitational Potential Energy at the Earth's Surface

The gravitational potential energy of a satellite of mass at the Earth's surface is:

$ = Gravitational constant

= Mass of the satellite

= Mass of the Earth

= Radius of the Earth

Step 2: Gravitational Potential Energy at the Orbit

The altitude of the orbit is , so the distance from the center of the Earth to the satellite is . The gravitational potential energy at this orbit is:

\) is given by:

\) $