The diagram shows point P is halfway between the centres of two equal spherical masses that are separated by a distance of 2 × 10 9 m Calculate the gravitational field strength at a point P and state the direction of the gravitational field strength at point Q .

Since the point is midway between the two identical masses, their effect will be the same. Since the gravitational force is always attractive, the force of these masses at the point P , will be equal and opposite. Therefore, they will balance each other and the net gravitational field strength (force per unit mass) will be zero. At the point Q , both masses have equal attractive force directed towards their centres, These forces can be resolved in two perpendicular directions. The horizontal force will be equal and in opposite directions, hence cancel out. The vertical components will be added up. Therefore, the direction of the gravitational field at the point Q will be vertically upwards

Assume that the force of gravity between two point masses is given by F = G m 1 m 2 r n , where n is a constant. Derive the law relating period to orbit radius for this force,

Since the necessary centripetal force for the orbital motion is provided by the gravitational force. Hence we can say that: G . m 1 m 2 r n = m 2 v 2 r or v = Gm 1 r n - 1 Since the velocity in terms of the time period is v= 2 π r T , Therefore we get Gm 1 r n - 1 = 2 πr T Squaring both sides and rearranging, we get, GM 1 r n - 1 = 4 π 2 r 2 T 2 or T 2 = 4 π 2 r n + 1 Gm 1

Assume that the force of gravity between two point masses is given by F = G m 1 m 2 r n , Deduce the value of n if this law is to be identical with Kepler's third law.

Since the recessionary centripetal force for the orbital motion is provided by the gravitational force. Hence, we can say that: G . m 1 m 2 r n = m 2 v 2 r or v = Gm 1 r n - 1 Since the velocity in terms of the time period is v= 2 π r T , Therefore we get Gm 1 r n - 1 = 2 πr T Squaring both sides and rearranging, we get, GM 1 r n - 1 = 4 π 2 r 2 T 2 or T 2 = 4 π 2 r n + 1 Gm 1 The Kepler's third law states that "square of the orbital time period of a planet is proportional to the cube of semi major axis of the orbit ( r 3 ) " Therefore, the term r n + 1 must be equal to r 3 in Kepler's relation. Hence r n + 1 = r 3 or n = 2

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The mass of the Moon is about $81$ times less than that of the Earth. Estimate the fraction of the distance from the Earth to the Moon where the gravitational field strength is zero. (Take into account the Earth and the Moon only.)

Important Questions on Circular Motion and Gravitation

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Diploma

IMPORTANT

Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 22

The diagram shows point $P$ is halfway between the centres of two equal spherical masses that are separated by a distance of $2 \times 10^{9} m$ Calculate the gravitational field strength at a point $P$ and state the direction of the gravitational field strength at point $Q$ .

Question Image

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 23

A satellite orbits the earth above the equator with a period equal to $24 hrs .$ Determine the height of the satellite above the earth.

[ $G = (6.67 m^{3} {kg}^{- 1} s^{- 2})$ ]

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Diploma

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 23

A satellite orbits the earth above the equator with a period equal to

24 hrs .

Suggest an advantage of such a satellite.

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Diploma

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 24

The Hubble Space Telescope is in the orbit around the earth at a height of $560 km$ above the earth's surface Calculate the Hubble's speed.

$[G = 6.67 m^{3} {kg}^{- 1} s^{- 2}]$

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Diploma

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 24

In a servicing mission, a space shuttle spotted the Hubble telescope a distance of $10 km$ . Estimate how long it took the shuttle to catch up with the Hubble, assuming that the Shuttle was moving in a circular orbit just $500 m$ below Hubble's orbit of $560 km$ .

$[G = 6.67 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2}]$

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Diploma

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 25

Assume that the force of gravity between two point masses is given by

F = G \frac{m_{1} m_{2}}{r^{n}}

, where

n

is a constant. Derive the law relating period to orbit radius for this force,

HARD

Diploma

IMPORTANT

Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Test yourself>Q 25

Assume that the force of gravity between two point masses is given by

F = G \frac{m_{1} m_{2}}{r^{n}}

, Deduce the value of

n

if this law is to be identical with Kepler's third law.

EASY

Diploma

IMPORTANT

Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 1

A child is sitting at the edge of a merry-go-round.The arrow shows the velocity of the child.At the instant shown, he releases the ball on to the ground.Which is the path of the ball according to a stationary observer on the ground ?

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The mass of the Moon is about 81 times less than that of the Earth. Estimate the fraction of the distance from the Earth to the Moon where the gravitational field strength is zero. (Take into account the Earth and the Moon only.)

Important Questions on Circular Motion and Gravitation

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The mass of the Moon is about $81$ times less than that of the Earth. Estimate the fraction of the distance from the Earth to the Moon where the gravitational field strength is zero. (Take into account the Earth and the Moon only.)