Chp 7: G-field

Question 1

Define gravitational field strength

Answer 1

Gravitational field strength in a gravitational field is the gravitational force per unit mass acting on a small test mass placed at that point

Question 2

Define gravitational potential

Answer 2

Gravitational potential at a point in a gravitational field is the work done per unit mass by an external agent in bringing a small test mass from infinity to that point without a change in kinetic energy

Question 3

Explain why the acc. of free fall near the earth’s surface is approximately constant.

Answer 3

Since the earth has an extermely large radius, gravitational field lines are approximately parallel and equally spaced.

Hence gravitational field strength is approximately constant.

By N2L, since object is free-falling,
Fnet = mg
ma = mg
a=g (diff units!!!)
a ~ constant

Question 4

Why we use a small test mass?

Answer 4

So as to not distort the field created by mass M, but only experience the field created by M.

Question 5

What is escape speed?

Answer 5

It is the minimum speed in which a body should be projected in order to escape a planet’s gravitational field.

Question 6

For a satellite to be in geostationary orbit, it must:

Answer 6

1. placed directly above the Earth’s equator
2. have an orbital period of 24 hours, so that its anngluar speed is the same as Earth’s
3. must move in the same direction as Earth (from west to east)

Question 7

Applications of geostationary satellite

Answer 7

1. communication satellites. they remain stationary above a point on Earth so no need to adjust the direction of the satellite dish to receive or trasmit signals to or fro from the geostationary satellite
2. meteorological applications and remote img. able to scan a large section of Earth continuously.

Question 8

Disadvantages of geostationary satellites

Answer 8

1. Due to its high altitude, signals need to take a longer time to reach the satellite, causing a lag time for live broadcast
2. Since they are positioned at a high altitude, the resolution of the images may not be good

Question 9

Explain why a person in a satellite experience ‘weightlessness’ although the gravitational field strength is non-zero.

Answer 9

As the satellite undergoes circular orbit, the gravitational force exerted by Earth on satellite just sufficiently provides for the centripetal force for the satellite to undergo circular motion [1]

The centripetal acceleration of the satellite and the man is the same. Therefore, since a=g, normal contact force = 0, the man will exp ‘weightlessness’.

Question 10

Explain why the satellite ‘burns’ up in the atmosphere when total energy of it decreases

Answer 10

As the satellite speeds up, it experiences greater drag. Since drag increases with speed, more energy dissipated as heat which eventually causes the satellite to ‘burn’ up.

Question 11

Explain why the normal reaction on the mass will have different values at the equator and poles

Answer 11

At the poles: N = Fg
At the equator: N = Fg -mrw2

For the mass at the equator, the centripetal force is provided by Fg - N cuz the planet is rotating. However at the poles, the mass is not rotating thus N=Fg.

Normal reaction force is larger at the poles compared to the equator

Question 12

What can cause variations in the acceleration of free fall at surface of planet?

Answer 12

1. planet may not be a perfect sphere. the different radii will affect the gravitational field strength depending on the different position of the mass.
2. planet may not have a uniform density

Question 13

Explain why the centripetal force acting on both stars has the same magnitude

Answer 13

The gravitational force between both stars provide for the centripetal force for the stars to orbit[1]

By N3L, the gravitational force exerted by A on B is equal in magnitude as the gravitational force exterted by B on A[1]

Question 14

Explain why gravitational potential is a negative quantity

Answer 14

At infinity, the gravitational potential is zero.

Gravitational force is attractive. To move a mass from infinity to a point in a field, the external force is opposite in direction to the displacement.

Hence work done by external force is negative.

Question 15

Suggest why the gas particles at the equator of the star are more likely to escape the surface than those at the poles

Answer 15

At the equator, it has the largest radius of rotation, whereas at the poles, the radius of rotation is zero.

Since v=rw, for constant w, the gas particles at the equator has the largest linear speed. Hence…

Question 16

Define Newton’s law of gravitation

Answer 16

It states that two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 17

Define gravitational field

Answer 17

is a region of space in which a mass experiences a gravitational force

Question 18

1 similarity & 1 diff: Electric potential vs gravitational potential

Answer 18

Similarity: Both inversely proportional to the distance of the point from the point source.

Diff: Electric potential values is + for + point charge or - for - point charge but gravitational potential values always -

Question 19

Can we use 𝑲𝑬 = 𝑮𝑴𝒎/𝟐𝒓 to calculate change in KE as a satellite is launched from the surface of Earth to orbit around the Earth?

Answer 19

Cannot.

On Earth’s surface, got Fg and N acts on satellite.

However, KE =GMm/2r was derived for an orbiting satellite where only Fg acts on satellite.

Question 20

Why Newton’s law of gravitation can calculate force between the stars even tho need to take their diameters into consideration?

Answer 20

Separation between 2 masses is&raquo_space;> diameters of the stars. So, they can be assumed to be point masses

Question 21

Why centripetal forces acting on 2 binary stars are equal in magnitude?

Answer 21

Fg between them provide Fc.

By N3L, Fg by A on B is equal and opp in direction of Fg by B on A. So…

Question 22

How to use light intensity to determine the period of binary stars?

Answer 22

As star orbits, at a certain time, A blocks B, so intensity detected of starts decreases -T1. T2 is next intensity decrease. Period = 2(T2 - T1)

Question 23

Why 2 stars revolving around their common centre of mass must have the same angular velocity?

Answer 23

to ensure Fg acting on them are collinear and pass through their common centre of rotation

Question 24

Even though the Fg on each star is directed towards each other, the stars do not move towards each other and collide with each other. Explain why the stars maintain a circular orbit.(pic shows 2 stars moving in orbit, having same m and v, at opp side of the orbit)

Answer 24

Fg by stars on each other is always perpendicular to their velocities. Fg just sufficient to cause centripetal acceleration of the centripetal acceleration of the stars

Question 25

Why the gravitational field strength at the moon has the same magnitude and direction as the centripetal acceleration of the moon.

Answer 25

• only force acting on moon is Fg by planet on moon.
• Fg provides Fc
• By N2L: Fg = Fc
• m(moon)g = m(moon)a(centripetal)
• a = g (same direction)

Question 26

A satellite that is orbiting Earth, has its orbital radius reduced. State and explain the
following effect on

GPE of the satellite:

KE of the satellite:

Answer 26

GPE of the satellite: Since GPE is − 𝐺𝑀𝑚/𝑟 , as the orbital radius is reduced, the GPE is more negative, hence GPE decreases.

KE of the satellite: Since KE = 𝐺𝑀𝑚/2𝑟 , as the orbital radius is reduced, the KE increases.

Question 27

Why geostationary orbits must be directly above the equator?

Answer 27

• to ensure object rotates about the same axis as Earth
• line joining object and centre of earth has to pass thru equator and perpendicular to earth’s axis of rotation
• if object lies above other latitudes besides the equator, its axis of rotation will not coincide w that of earth’s.
  -so, will not be geostationary

Question 28

Why acceleration of free fall at the equator is smaller than the acceleration of free fall at the pole. (when man is standing on weighing scale)

Answer 28

By N3L:

Nby man on weighing scale = Nby weighing scale on man

Equator:
Fnet on man = mg - Nequator
Fnet on man provides Fc
Fnet on man = Fc = mrw2
Nequator = mg - mrw2

Pole:
r=0
Npole = mg - m(0)w2
Npole = mg

Since Nequator < Npole, Npole = mg = ma, aequator < apole

Question 29

Why it is easier to project an object from the Moon’s surface to the Earth compared to projecting an object from Earth’s surface to the moon. (draw graph for better visual)

Answer 29

To project an object from Earth to just reach Moon, the initial KE required must be sufficient to just reach the null point (gnet = 0). Similarly, to project an object from Moon to just reach Earth, the initial KE required must be just sufficient to reach the null point too. However, the increase in GPE from Earth to null point is more than the increase in GPE from Moon to null point, hence greater initial KE required to go from Earth to Moon.

Question 30

As the satellite orbits the Earth, it gradually loses energy because of air resistance. State and explain the effect of this change on the radius of the orbit and the speed of the satellite.

Answer 30

TE = KE + GPE
Fg provides Fc
… until you find KE = Gmm/2r
TE = -½(Gmm/r)

When satellite lose energy, TE more negative, r decreases, KE increases, v increases.