
The escape velocity from the earth is about 11 km / second. The escape velocity from a planet having twice the radius and the same mean density as the earth, is:
(A) 22 km/sec
(B) 11km/sec
(C) 5.5km/sec
(D) 15.5km/sec
Answer
139.5k+ views
Hint: To answer this question we should first begin with the formula for the escape velocity. Then it is required to find the escape velocity of earth and the planet. After the values are mentioned in the equation we have to evaluate them to get the required answer.
Complete step by step answer:
We should know that the escape velocity is denoted by ${V_{es}}$. The formula of the escape velocity is given by:
${V_{es}} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt {\dfrac{{2G}}{R}.\dfrac{4}{3}\pi {R^3}\rho } $
Here R denotes the radius of the earth, G denotes the universal gravitational constant and M is the mass. So we can write the value of ${V_{es}}$ as:
$
{V_{es}} = \sqrt {\dfrac{8}{3}\pi {G_\rho }{R^2}} = R\sqrt {\dfrac{{8\pi {G_p}}}{3}} \\
\Rightarrow {V_{es}}\propto R \\
$
So we can write that:
$\dfrac{{{{({V_{es}})}_{planet}}}}{{{{({V_{es}})}_{earth}}}} = \dfrac{{2R}}{R} = 2$$\dfrac{{{{({V_{es}})}_{planet}}}}{{{{({V_{es}})}_{earth}}}} = \dfrac{{2R}}{R} = 2$
Now we have to evaluate the above expression:
$
\Rightarrow {({V_{es}})_{planet}} = 2 \times {({V_{es}})_{earth}} \\
\\
$
We have to put the values in above expression to get the value as:
${({V_{es}})_{planet}} = (2 \times 11)km/\sec = 22km/\sec $
So we can say that the escape velocity from a planet having twice the radius and the same mean density as the earth, is 22 km/sec.
Hence the correct answer is option A.
Note: We should know that the escape velocity is defined as the minimum amount of energy that is required for any free or we can say non- propelled object to free itself from the gravitational influence of a huge body. This signifies that the object will want to move away at a distance which is infinite from the huge structure. It should be known to us that more will be the speed, the escape velocity will have a positive speed at the infinity. Even if not mentioned we should always consider the escape velocity of earth to be 11 km/ sec.
Complete step by step answer:
We should know that the escape velocity is denoted by ${V_{es}}$. The formula of the escape velocity is given by:
${V_{es}} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt {\dfrac{{2G}}{R}.\dfrac{4}{3}\pi {R^3}\rho } $
Here R denotes the radius of the earth, G denotes the universal gravitational constant and M is the mass. So we can write the value of ${V_{es}}$ as:
$
{V_{es}} = \sqrt {\dfrac{8}{3}\pi {G_\rho }{R^2}} = R\sqrt {\dfrac{{8\pi {G_p}}}{3}} \\
\Rightarrow {V_{es}}\propto R \\
$
So we can write that:
$\dfrac{{{{({V_{es}})}_{planet}}}}{{{{({V_{es}})}_{earth}}}} = \dfrac{{2R}}{R} = 2$$\dfrac{{{{({V_{es}})}_{planet}}}}{{{{({V_{es}})}_{earth}}}} = \dfrac{{2R}}{R} = 2$
Now we have to evaluate the above expression:
$
\Rightarrow {({V_{es}})_{planet}} = 2 \times {({V_{es}})_{earth}} \\
\\
$
We have to put the values in above expression to get the value as:
${({V_{es}})_{planet}} = (2 \times 11)km/\sec = 22km/\sec $
So we can say that the escape velocity from a planet having twice the radius and the same mean density as the earth, is 22 km/sec.
Hence the correct answer is option A.
Note: We should know that the escape velocity is defined as the minimum amount of energy that is required for any free or we can say non- propelled object to free itself from the gravitational influence of a huge body. This signifies that the object will want to move away at a distance which is infinite from the huge structure. It should be known to us that more will be the speed, the escape velocity will have a positive speed at the infinity. Even if not mentioned we should always consider the escape velocity of earth to be 11 km/ sec.
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