so that it goes beyond the gravitational field of the earth, then the velocity is said to escape velocity. So the escape velocity is defined as the velocity which will take a body to go the infinite distance away above the surface of the earth when projected upwards.
Suppose that the earth is a perfect sphere of radius R having mass M. let a body of mass m is to be projected from a point A on the earth’s surface as shown in the figure. Join OA and produce it further. Let us take two points P and Q which are at the distance x and (a + dx) from the centre of the earth.
To calculate the escape velocity of the earth, let the minimum velocity to escape from the earth’s surface is v_{e}. Then, kinetic energy of the object of mass m is
$$K.E= \frac{1}{2}mV_e^2$$ When the projected object is at point P which is at a distance x from, the centre of the earth, the force of gravity between the object and earth is\begin{align*} \\ F &= \frac {GMm}{x^2} \\ \end{align*}Work done is taking the body against gravitational attraction from P to Q is given by \begin{align*}\\ dW &= F dx = \frac {GMM}{x^2}\\ \end{align*}
The total amount of work done in taking the body against gravitational attraction from the surface of the earth to infinity can be calculated by integrating the above equation within the limits x=R to x = \(\infty \), hence total work done is
w = \(\int_{R}^{\infty}\) dw =\(\int_{R}^{\infty}\) \(\frac{GMm}{x^2}\)dx
=GMm\(\int_{R}^{\infty}\)x^{-2}dx =GMm [\(\frac{X^{-1}}{-1}]_{R}^{\infty}\) =GMm [\(\frac{1}{x}]_{R}^{\infty}\) = -GMm [\(\frac{1}{\infty}\) -\(\frac{1}{R}\)]
or W = \(\frac{GMm}{R}\)
To escape the object from theearth’s surface, kinetic energy given must be equal to the work done against gravity going from the earth’s surface to inifinity,
\begin{align*} K.E. &=W \\ \text {or,} \frac 12 mv_e^2 &= \frac {GMm}{R} \\ v_e = \sqrt {\frac {2GM}{R} } \\ g &= \frac {GM}{R^2} \\ \therefore v_e &= \sqrt {2gR} \\ \end{align*}
The relation shows that the escape velocity of an object does not depend on the mass of the projected object but only on the mass and radius of the planet from which it is projected. For the earth, g= 9.8 ms^{-2} and R = 6.4×10^{6} m, then
\begin{align*} \text {escape velocity of the earth,} v_e &= \sqrt {2\times 9.8 \times 6.4 \times 10^6 } \\ &= 11.2 \times 10^3 m/s = 11.2 km/s. \end{align*}
c. If a body mass m is lying at a location where the effective acceleration due to gravity is g’, then the effective weight of the body = mg’. The effective weight of the body will be zero is g’ = 0.
In the following circumstances, the body will be in weightlessness condition.
Let us suppose a very high tower on the surface of the earth. Consider a body is projected horizontally from the top of the tower with some velocity. The body will follow a parabolic path under the effect of gravity and strikes the earth’s surface at point A_{1} as shown in the figure. If the body is projected with the velocity greater than the initial velocity, then the body will strike the surface at point A_{2} which is farther from A_{1}. If the velocity is gradually increased, the horizontal range will also increase and finally a stage will come when the body will not strike the earth’s surface, but will always be in a state of free fall under gravity in an attempt to fall to the earth. Then the body will describe a stable circular path around the earth and becomes a satellite of the earth. The minimum velocity required to put the satellite into its orbit around the earth is called orbital velocity of the satellite.
\begin{align*} \text {escape velocity of the earth,} v_e &= \sqrt {2\times 9.8 \times 6.4 \times 10^6 } \\ &= 11.2 \times 10^3 m/s = 11.2 km/s. \end{align*}
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edil
what is the use of weightlessness
Mar 09, 2017
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edil
what is the use weightlessness
Mar 09, 2017
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