Prove that the total energy of earth (mass m ) as it orbits the Sum (mass M ) can be expressed as either E = - 1 2 mv 2 or E = - GMm 2 r where r is the radius of Earths' circular orbit. Calculate the energy numerically.

The kinetic energy of a moving body is given by E = 1 2 mv 2 ............................................................(1) where m = mass of the body and v = its speed. Now we can consider the earth as the satellite orbiting at a height equal to the radius of the orbit around the sun (= r ) The orbital velocity of the earth around the sun is v = GM r ....................(2) Substituting in the equation (1), we get E = - 1 2 m GM r 2 = - GMm 2 r Negative energy means It binds the two. Putting the standard values in the formula we have E = - G M m 2 ( r ) = - 6 . 67 × 10 - 11 N k g - 2 m - 2 × 2 × 10 30 k g × 5 . 97 × 10 24 k g 2 × ( 1 . 5 × 10 11 m + 6 . 4 × 10 6 + 6 . 96 × 10 8 ) where r = distance of the centres of the earth and the sun. E = - 5 . 3 × 10 33 J

Prove that the total energy of earth (mass m ) as it orbits the Sum (mass M ) can be expressed as either E = - 1 2 mv 2 or E = - GMm 2 r where r is the radius of Earths' circular orbit. Calculate the energy numerically.

The kinetic energy of a moving body is given by E = 1 2 mv 2 ............................................................(1) where m = mass of the body and v = its speed. Now we can consider the earth as the satellite orbiting at a height equal to the radius of the orbit around the sun (= r ) The orbital velocity of the earth around the sun is v = GM r ....................(2) Substituting in the equation (1), we get E = - 1 2 m GM r 2 = - GMm 2 r Negative energy means It binds the two. Putting the standard values in the formula we have E = - G M m 2 ( r ) = - 6 . 67 × 10 - 11 N k g - 2 m - 2 × 2 × 10 30 k g × 5 . 97 × 10 24 k g 2 × ( 1 . 5 × 10 11 m + 6 . 4 × 10 6 + 6 . 96 × 10 8 ) where r = distance of the centres of the earth and the sun. E = - 5 . 3 × 10 33 J

The total energy of satellite during launch from the earth surface is E = - GMm 5 r , where R is the radius of the earth. Explain why this satellite will not escape the earth

Given the energy of the satellite = E = - GMm 5 r The satellite will not escape because the total energy is negative. This will bound the satellite with the earth. The kinetic energy of the satellite will be K = + GMm 5 r = 1 2 mv 2 or v = 2 GM 5 r This velocity is less than the escape velocity which is v e = GM r . Therefore the satellite will not escape from the earth.

Show that the escape velocity from the surface of a planet of radius R can be written as V esc = 2 gR where g is the gravitational field strength on the planet's surface.

The escape velocity is defined as the minimum speed required by a body to project it vertically upward from the surface of the planet so that it never returns. Since the work done to displace the body to the infinity from the surface of the planet is the potential energy at infinity, so we have, the potential energy of the mass m at infinity, U = - GMm R ................................................(1) The kinetic energy of the body is given by K . E . = 1 2 mv e 2 where v e = escape velocity ............(2) The body will escape from the gravitational pull if the KE = magnitude of PE at ∞ . Equating the equations we get 1 2 mv e 2 = - GmM R v e = 2 GM R GM R 2 = g or GM = g R 2 , we can write, v e = 2 gR

E M B I B E

Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 28

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Prove that the total energy of earth (mass $m$ ) as it orbits the Sum (mass $M$ ) can be expressed as either $E = - \frac{1}{2} {mv}^{2} or E = - \frac{GMm}{2 r}$ where $r$ is the radius of Earths' circular orbit. Calculate the energy numerically.

Important Questions on Fields (HL)

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Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 29

The diagram shows two identical satellites in circular orbits. Which satellite has the larger kinetic emery? Question Image

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Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 29

The diagram shows two identical satellites in circular orbits. Which satellite has the larger potential emery? Question Image

HARD

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IMPORTANT

Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 29

The diagram shows two identical satellites in circular orbits. Which satellite has the larger total energy ? Question Image

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Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 30

The total energy of satellite during launch from the earth surface is

E = - \frac{GMm}{5 r}

, where

R

is the radius of the earth. Explain why this satellite will not escape the earth

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Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 30

The total energy of satellite during launch from the earth surface is

E = - \frac{GMm}{5 r}

, where

r

is the radius of the earth. The satellite eventually settles into a circular orbit. Calculate the radius of the orbit in terms of

r

.

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Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 31

A satellite is in a circular orbit around the earth. The satellite turns its engine and the satellite now find itself in a new circular orbit larger radius. State and explain whether the work done by the engine positive, zero or negative.

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Physics for the IB Diploma 6th Edition>Chapter 10 - Fields (HL)>Test yourself>Q 32

The diagram shows a planet orbiting the Sun. Explain why at a point $A$ and $P$ of the orbit the potential energy of the planet assume its maximum and minimum values and determine which is which. Hence, determine at what point of the orbit, the planet has the greatest speed.

Question Image