What is the escape velocity from Jupiter Given that the mass is 300 times that of the Earths and its radius is 10times larger?

The escape velocity formula is independent of the properties of the escaping object. The only thing that matters is the mass and radius of the celestial body in question:

vₑ = √(2GM/R)

M is the mass of the planet, R is its radius, and G is the gravitational constant. It is equal to G = 6.674 × 10-11 N·m²/kg².

The formula for escape velocity, also known as the second cosmic velocity, is derived directly from the law of conservation of energy. At the moment of launch, the object has some potential energy PE and some kinetic energy KE. The energy at launch LE can be hence presented as follows:

PE + KE = -GMm/R + ½mv²

where m is the mass of the starting object, and v is the escape velocity.

When the object finally escapes, it is located so far from the planet that its potential energy is equal to zero. Also, it can have virtually no speed, so its kinetic energy is also equal to zero. That means that the total final energy is equal to:

PE + KE = 0 + 0 = 0

Because the total energy must be conserved, it means that the initial energy is also equal to zero. Simplifying the first equation, we get:

0 = -GMm/R + ½mv²

v = √(2GM/R)

Check our kinetic energy calculator and potential energy calculator for more details on the topic of energy.

Answer

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Hint: We have to find out the expression for the escape velocity of Jupiter and Earth. Then we have to equate the expressions after mentioning the values from the question. This will give us the value of the escape velocity of Jupiter, which is the required answer.

Complete step by step answer:

We know that it is given that:${M_J} = 318{M_e}$ and${R_J} = 11.2{R_e}$${V_e}$= 11.2 km/sHere ${M_J}$ is the mass of Jupiter and ${M_e}$ is the mass of Earth. ${R_J}$ denotes Jupiter's radius and ${R_e}$ is Earth’s radius. ${V_e}$ is the escape velocity from the Earth’s surface. So we know that now:${V_J} = \sqrt {\dfrac{{2G{M_J}}}{{{R_J}}}} $and${V_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} $G refers to the universal gravitational constant. So we can write the expression for the escape velocity of Jupiter as follows:$ \Rightarrow {V_J} = {V_e}\sqrt {\dfrac{{{M_J}}}{{{M_e}}} \times \dfrac{{{R_e}}}{{{R_J}}}} $Putting the values in the above equation we get:$ {V_J} = 11.2 = {\left\{ {\dfrac{{318{M_e}}}{{{M_e}}} \times \dfrac{{{R_e}}}{{11.2{R_e}}}} \right\}^{\dfrac{1}{2}}} \\ \Rightarrow {V_J} = 11.2{\left( {\dfrac{{318}}{{11.2}}} \right)^{\dfrac{1}{2}}} = 59.7km/s \\ $

Hence we can say that the escape velocity of a body from Jupiter’s surface is 59.7 km/s.

Note: We should know the definition of escape velocity for a better understanding. So we define escape velocity as the speed at which a body must travel so as to break itself free from a planet’s gravitational pull or moon’s gravitational pull. The body then enters into an orbit. It is said to be the minimum lowest velocity required by the body to overcome the gravitational pull.

The unit of escape velocity is m/s. For solving problems related to the escape velocity we should remember that to obtain the escape velocity we need to multiply the altitude with the square root of 2. Then we will drive the velocity that will be required to escape the orbit and the gravitational field which is controlling the orbit.

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We don’t have your requested question, but here is a suggested video that might help.

The planet Jupiter is more than 300 times as massive as Earth, so it might seem that a body on the surface of Jupiter would weigh 300 times as much as on Earth. But it so happens that a body would weigh scarcely 3 times as much on the surface of Jupiter as on the surface of Earth. Discuss why this is so, using the terms in the equation for gravitational force to guide your thinking.

A rocket is fired vertically with a speed of 5 km/s from the earth’s surface. How far from the earth’s surface does the rocket go before returning to the earth? Mass of the earth = 6.0 x 1024 kg; mean radius of the earth = 6.4 x 106 m; G =6.67 x 10-11 N m2 kg-2.
Acceleration due to gravity at the surface of the earth = 9.8 m/s2.

When a body is thrown a body in the upward direction, the kinetic energy of the body converts into potential energy. Potential energy of the body at ground is,

If the body is raised to a height h from the earth, then potential energy at height h is,

Therefore increase in potential energy is,

The increase in potential energy is at the cost of kinetic energy, therefore

Here, we have

Therefore,

That is, the rocket travels a distance of 1.6 x 106 m before returning to the Earth's surface.