Grav Field

Question 1

Define Newton’s law of gravitation

Answer 1

Two points masses attract each other with a force that is directly proportional to the product of their point masses and inversely proportional to the square of the distance between them

Question 2

Grav force formula

Answer 2

GMm/r^2

Question 3

Define Gravitational field

Answer 3

a region of space in which a mass experiences a force due to the presence of another mass

Question 4

Define gravitational field strength at a point in the gravitational field

Answer 4

The gravitational force per unit mass acting on a small test mass placed at that point

Question 5

Gravitational field strength formula

Answer 5

(GM)/r^2

Question 6

Define Gravitational potential energy

Answer 6

The work done by a test mass in bringing a mass from infinity to that point

Question 7

Gravitational potential energy formula

Answer 7

U = -(GMm)/r

Question 8

Define gravitational potential

Answer 8

The work done per unit mass by an external agent in bringing a test mass from infinity to that point without a change in kinetic energy.

Question 9

Gravitational potential formula

Answer 9

-(GM)/r

Question 10

Kepler’s third law

Answer 10

The square of the period of revolution of the planets is directly proportional to the cubes of the mean distances from the sun

Question 11

Define gravitational force

Answer 11

The force acting on a mass that is placed in a gravitational field

Question 12

Explain why for changes in vertical position of a point mass near the earth’s surface, gravitational field strength may be considered to be constant

Answer 12

Changes in height near the earth’s surface is much lesser than the radius of the earth,
so gravitational field lines are almost parallel and equally spaced to one another near the earth’s surface,
hence grav field strength is constant.

Question 13

Explain why, at the surface of a planet, g field strength is numerically equal to the acceleration of free fall. (Relate to grav force)

Answer 13

At the surface of a planet, a mass will experience a grav force equal to weight mg, where g is the g field strength at the surface.
By N2L, resultant force acting on the mass = grav force so g = a

Question 14

Explain what it means for a mass to feel ‘weightless’.

Answer 14

Weightless -> normal contact force is zero. In this situation, only force that provides the centripetal force is his gravitational force which is also equal in magnitude to the centripetal force.
By N2L, the acceleration of the mass is numerically equal to g at that point, hence apparent weight of astronaut is zero.

mg - N = ma, and mg = ma, so N = 0

Question 15

Suggest quantitatively why it may be assumed that the sun is isolated in space from other stars.

Answer 15

Quantitatively, the g force of attraction between the sun and star causes a very small hence negligible attraction on the sun due to the sun’s large mass.

There are many other stars around the sun so the net grav force on the sun is 0.

Question 16

Explain why the centripetal force on two stars in binary orbit has the same magnitude.

Answer 16

Grav force provides centripetal force. By N3L, grav force on each star is equal in magnitude and opposite in direction, so centripetal force will have the same magnitude.

Question 17

Explain why binary stars must orbit with the same angular velocity.

Answer 17

Grav force provides centripetal force which acts along the line joining the centre of the stars. So stars must always be on opposite ends of the line through their common centre of the mass, and thus must orbit with the same angular velocity.

Question 18

Why is grav potential near an isolated mass always negative?/a negative quantity?

Answer 18

Grav force is attractive in nature so the external agent has to exert a force equal and opposite such that the test mass does not experience a change in kinetic energy.
Since the external agent is opposite to the displacement of the test mass, the work done per unit mass by the external agent is negative

Question 19

What is a geostationary orbit

Answer 19

1. A circular orbit in the earth’s equatorial plane
2. An orbital period of 24 hours
3. The same direction of rotation as the earth from west to east.

Question 20

One advantage and one disadvantage of the use of geostationary satellites

Answer 20

Advantage: provides continuous surveillance if the region under it

Disadvantage: unable to provide coverage for the polar regions

Question 21

Explain why a geostationary satellite must be positioned directly above the equator.

Answer 21

This is to ensure the satellite rotates about the same axis as the earth. The line joining the object and the centre of the earth has to pass through the equator and rests along the plane perpendicular to the earth’s axis of rotation.

If the object lies above other latitudes besides the equator, it’s axis of rotation would not coincide with that of the earth’s. as such it would be above different latitudes as it rotated and will not be geostationary.

Question 22

Using Newton’s laws of motion, explain why the two planets orbit about the center of mass of the double planet system.

Answer 22

Each planet orbits due to the gravitational force exerted by the other. As a system, there are no external forces exerted on them.
Thus by N1L, the center of mass must remain stationary. So the center of mass is the center of both planet’s orbit.

Question 23

Derive grav field strength

Answer 23

ΣF = ma
(GMm/R²) = ma

a = GM/R

Question 24

Derive Gravitational potential energy using terms

Answer 24

Integrate GMm/r²

Question 25

Derive escape velocity.

Answer 25

E at Earth’s surface = E at infinity
(Ep + Ek) Earth’s surface = (Ep + Ek) at infinity

-(GMem)/Re + ½ m[v(esc)]² =0+0
½ m[v(esc)]² = GMe/Re
v(esc) = (2GMe/Re)½

Question 26

Derive the formula for a mass of a planet in terms of r, G and T

Answer 26

ΣF = ma
GMem/R² = mrω²
GMem/R² = mr(2π/T)²
Me = 4πr³/GT²

Question 27

Derive kinetic energy of a satelite

Answer 27

ΣF = ma
Gmm/r² = mv²/r
½mv² = GMm/2r