# 0.11 Projection in gravitational field  (Page 2/5)

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$⇒\Delta U=-\frac{GMm}{\left(R+h\right)}-\left(-\frac{GMm}{R}\right)$

$⇒\Delta U=GMm\left(\frac{1}{R}-\frac{1}{R+h}\right)$

We can eliminate reference to gravitational constant and mass of Earth by using relation of gravitational acceleration at Earth’s surface ( $g={g}_{0}$ ),

$g=\frac{GM}{{R}^{2}}$

$⇒GM=g{R}^{2}$

Substituting in the equation of change in potential energy, we have :

$⇒\Delta U=mg{R}^{2}\left(\frac{1}{R}-\frac{1}{R+h}\right)$

$⇒\Delta U=mg{R}^{2}X\frac{h}{R\left(R+h\right)}$

$⇒\Delta U=\frac{mgh}{1+\frac{h}{R}}$

It is expected that this general formulation for the change in potential energy should be reduced to approximate form. For h<<R, we can neglect “h/R” term and,

$⇒\Delta U=mgh$

## Maximum height

For velocity less than escape velocity (the velocity at which projectile escapes the gravitation field of Earth), the projected particle reaches a maximum height and then returns to the surface of Earth.

When we consider that acceleration due to gravity is constant near Earth’s surface, then applying conservation of mechanical energy yields :

$\frac{1}{2}m{v}^{2}+0=0+mgh$

$⇒h=\frac{{v}^{2}}{2g}$

However, we have seen that “mgh” is not true measure of change in potential energy. Like in the case of change in potential energy, we come around the problem of variable acceleration by applying conservation of mechanical energy with reference to infinity.

${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$

$⇒-\frac{GMm}{R}+\frac{1}{2}m{v}^{2}=0+-\frac{GMm}{R+h}$

$⇒\frac{GM}{R+h}=\frac{GM}{R}-\frac{{v}^{2}}{2}$

$⇒R+h=\frac{GM}{\frac{GM}{R}-\frac{{v}^{2}}{2}}$

$⇒h=\frac{GM}{\frac{GM}{R}-\frac{{v}^{2}}{2}}-R$

$⇒h=\frac{GM-GM+\frac{{v}^{2}R}{2}}{\frac{GM}{R}-\frac{{v}^{2}}{2}}$

$⇒h=\frac{{v}^{2}R}{\frac{2GM}{R}-{v}^{2}}$

We can also write the expression of maximum height in terms of acceleration at Earth’s surface using the relation :

$⇒GM=g{R}^{2}$

Substituting in the equation and rearranging,

$⇒h=\frac{{v}^{2}}{2g-\frac{{v}^{2}}{R}}$

This is the maximum height attained by a projection, which is thrown up from the surface of Earth.

## Example

Problem 1: A particle is projected vertically at 5 km/s from the surface Earth. Find the maximum height attained by the particle. Given, radius of Earth = 6400 km and g = 10 $m/{s}^{2}$ .

Solution : We note here that velocity of projectile is less than escape velocity 11.2 km/s. The maximum height attained by the particle is given by:

$h=\frac{{v}^{2}}{2g-\frac{{v}^{2}}{R}}$

Putting values,

$⇒h=\frac{{\left(5X{10}^{3}\right)}^{2}}{2X10-\frac{{\left(5X{10}^{3}\right)}^{2}}{\left(6.4X{10}^{6}\right)}}$

$⇒h=\frac{\left(25X{10}^{6}\right)}{2X10-\frac{\left(25X{10}^{6}\right)}{\left(6.4X{10}^{6}\right)}}$

$⇒h=1.55x{10}^{6}=1550000\phantom{\rule{1em}{0ex}}m=1550\phantom{\rule{1em}{0ex}}km$

It would be interesting to compare the result, if we consider acceleration to be constant. The height attained is :

$h=\frac{{v}^{2}}{2}g=25X\frac{{10}^{6}}{20}=1.25X{10}^{6}=1250000\phantom{\rule{1em}{0ex}}m=1250\phantom{\rule{1em}{0ex}}km$

As we can see, approximation of constant acceleration due to gravity, results in huge discrepancy in the result.

## Escape velocity

In general, when a body is projected up, it returns to Earth after achieving a certain height. The height of the vertical flight depends on the speed of projection. Greater the initial velocity greater is the height attained.

Here, we seek to know the velocity of projection for which body does not return to Earth. In other words, the body escapes the gravitational influence of Earth and moves into interstellar space. We can know this velocity in verities of ways. The methods are equivalent, but intuitive in approach. Hence, we shall present here all such considerations :

## 1: binding energy :

Gravitational binding energy represents the energy required to eject a body out of the influence of a gravitational field. It is equal to the energy of the system, but opposite in sign. In the absence of friction, this energy is the mechanical energy (sum of potential and kinetic energy) in gravitational field.

#### Questions & Answers

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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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what is the actual application of fullerenes nowadays?
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Graphene has a hexagonal structure
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