The mass of Jupiter is 318 times that of earth and its radius is 11.2 times the earth’s radius. Estimate the escape velocity of a body from Jupiter’s surface given the escape velocity from the earth’s surface as 11.2 km/s.
Answer
200.7k+ views
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/s
Here ${M_J}$ 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.
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/s
Here ${M_J}$ 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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