
If the earth were made of wood, the mass of wooden earth is \[10\% \] as much as it is now without the change in its diameter. Calculate the escape velocity of space shot from the surface of the wooden earth. Radius of earth: R = 6400km
Mass of the earth: \[M = 6 \times {10^{24}}kg\]
Constant of gravitation: \[G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\]
Answer
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Hint Escape velocity is defined as the minimum velocity that a body must possess in order to escape the gravitational field of a planet. By comparing the potential energy of a body at height H above the surface of the planet with the kinetic energy of the body, we can derive the expression for escape velocity. Here we know all the quantities like G, M and R and hence we will substitute these in ${v_e} = \sqrt {\dfrac{{2GM}}{R}}$ to obtain escape velocity \[{v_e}\] .
Complete step by step answer:
The minimum velocity with which a body must be projected to enable it to just overcome the gravitational pull is known as escape velocity. It is given by,
${v_e} = \sqrt {\dfrac{{2GM}}{R}} $
Where,
G is the gravitational constant
M is the mass of the earth
R is the radius of the earth
Give: R = 6400km = \[\;6400 \times {10^3}\] m
\[G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\]
\[M = 10\% \times 6 \times {10^{24}}kg\]
\[M = 0.6 \times {10^{24}}kg\]
Substitute in the expression
$
{v_e} = \sqrt {\dfrac{{2 \times 6.67 \times {{10}^{ - 11}} \times 0.6 \times {{10}^{24}}}}{{6400 \times {{10}^3}}}} \\
\Rightarrow {v_e} = \sqrt {12.531 \times {{10}^6}} \\
\Rightarrow {v_e} = 3.54 \times {10^3}m{s^{ - 1}} \\
$
Hence the escape velocity if the earth is made of wood is \[3.54 \times {10^3}m{s^{ - 1}}\] .
Note:
The escape velocity for the body on planet earth under its normal condition is 11km/s. This means that a body must be thrown with a velocity of 11km/s to escape the gravitational field of the earth.
If the body is projected with a velocity less than escape velocity it may orbit around the earth or fall down. If the escape velocity of a body is equal to that of light then nothing can escape not even light. Such bodies are called black holes.
Complete step by step answer:
The minimum velocity with which a body must be projected to enable it to just overcome the gravitational pull is known as escape velocity. It is given by,
${v_e} = \sqrt {\dfrac{{2GM}}{R}} $
Where,
G is the gravitational constant
M is the mass of the earth
R is the radius of the earth
Give: R = 6400km = \[\;6400 \times {10^3}\] m
\[G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\]
\[M = 10\% \times 6 \times {10^{24}}kg\]
\[M = 0.6 \times {10^{24}}kg\]
Substitute in the expression
$
{v_e} = \sqrt {\dfrac{{2 \times 6.67 \times {{10}^{ - 11}} \times 0.6 \times {{10}^{24}}}}{{6400 \times {{10}^3}}}} \\
\Rightarrow {v_e} = \sqrt {12.531 \times {{10}^6}} \\
\Rightarrow {v_e} = 3.54 \times {10^3}m{s^{ - 1}} \\
$
Hence the escape velocity if the earth is made of wood is \[3.54 \times {10^3}m{s^{ - 1}}\] .
Note:
The escape velocity for the body on planet earth under its normal condition is 11km/s. This means that a body must be thrown with a velocity of 11km/s to escape the gravitational field of the earth.
If the body is projected with a velocity less than escape velocity it may orbit around the earth or fall down. If the escape velocity of a body is equal to that of light then nothing can escape not even light. Such bodies are called black holes.
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