
The escape velocity from the surface of the earth is (where ${R_E}$ is the radius of the earth)
A. $\sqrt {2g{R_E}} $
B. $\sqrt {g{R_E}} $
C. $\sqrt[2]{{g{R_E}}}$
D. $\sqrt {3g{R_E}} $
Answer
501.9k+ views
Hint: The minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a massive body, that is, to ultimately reach an infinite distance from it, is known as escape velocity in physics (specifically, celestial mechanics).
Formula used:
$V = \sqrt {\dfrac{{2GM}}{R}} $
Where, the escape velocity is $V$.$G$ stands for gravitational constant. The mass of the earth($M$) is \[6.67408{\text{ }}{10^{ - 11}}{\text{ }}{m^3}{\text{ }}k{g^{ - 1}}{\text{ }}{s^{ - 2}}{\text{ }}.\] $R$ is the distance from the gravitational source.
Complete step by step answer:
The minimum velocity at which a body must be projected to transcend the earth's gravitational force is called escape velocity. It is the speed at which an object must travel to escape the gravitational field, i.e., to leave the land without ever falling down. An object with this velocity at the earth's surface will completely escape the gravitational field of the planet, even though losses due to the atmosphere are taken into account.Escape velocity formula is given;
$V = \sqrt {\dfrac{{2GM}}{R}} $
An alternative expression for the escape velocity particularly useful at the surface on the body is
${V_{esc}} = g{R^2}$
Where $g$ denotes the acceleration caused by the earth's gravity.
Hence, Escape velocity is also given by
${V_{esc}} = \sqrt {2gR} $
Now according to the question;
The escape velocity from the surface of the earth is
${V_e} = \sqrt {\dfrac{{2GM}}{{{R_E}}}} $ --- (i)
Acceleration due to gravity
$g = \dfrac{{GM}}{{{R_E}^2}} \\$
$\Rightarrow \dfrac{{GM}}{{{R_E}}} = g{R_E}$--- (ii)
Substituting equation (ii) in equation (i) we get
$\therefore {V_e} = \sqrt {2g{R_E}} $
Hence, the correct option is A.
Note: The mass of the body, as well as the direction of projection of the body, have no bearing on escape velocity. The mass and radius of the planet or Earth from which the body is to be projected are the only factors.
Formula used:
$V = \sqrt {\dfrac{{2GM}}{R}} $
Where, the escape velocity is $V$.$G$ stands for gravitational constant. The mass of the earth($M$) is \[6.67408{\text{ }}{10^{ - 11}}{\text{ }}{m^3}{\text{ }}k{g^{ - 1}}{\text{ }}{s^{ - 2}}{\text{ }}.\] $R$ is the distance from the gravitational source.
Complete step by step answer:
The minimum velocity at which a body must be projected to transcend the earth's gravitational force is called escape velocity. It is the speed at which an object must travel to escape the gravitational field, i.e., to leave the land without ever falling down. An object with this velocity at the earth's surface will completely escape the gravitational field of the planet, even though losses due to the atmosphere are taken into account.Escape velocity formula is given;
$V = \sqrt {\dfrac{{2GM}}{R}} $
An alternative expression for the escape velocity particularly useful at the surface on the body is
${V_{esc}} = g{R^2}$
Where $g$ denotes the acceleration caused by the earth's gravity.
Hence, Escape velocity is also given by
${V_{esc}} = \sqrt {2gR} $
Now according to the question;
The escape velocity from the surface of the earth is
${V_e} = \sqrt {\dfrac{{2GM}}{{{R_E}}}} $ --- (i)
Acceleration due to gravity
$g = \dfrac{{GM}}{{{R_E}^2}} \\$
$\Rightarrow \dfrac{{GM}}{{{R_E}}} = g{R_E}$--- (ii)
Substituting equation (ii) in equation (i) we get
$\therefore {V_e} = \sqrt {2g{R_E}} $
Hence, the correct option is A.
Note: The mass of the body, as well as the direction of projection of the body, have no bearing on escape velocity. The mass and radius of the planet or Earth from which the body is to be projected are the only factors.
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