Grav Field

Question 1

Newton’s Law of gravitation

Answer 1

States that 2 pt masses attract each other with a force that is directly proportional to the product of their masses and inversely prop to the sq of the dist b/w them

Question 2

A grav field

Answer 2

Is a region of space in which a mass placed in that region experiences a grav force

Question 3

A grav field strength
def
symbol
eqn
vector/scalar

Answer 3

g, grav force exp per unit mass at that pt
g= - GM/r^2
vector

Question 4

Eqn of grav force
& symbol
vector/scalar

Answer 4

F=GMm/r^2

vector

Question 5

Grav potential energy
symbol
eqn
vector/scalar

Answer 5

U, Wk done by an external force in bringing the mass from infinity to that pt
U=-GMm/r
scalar

Question 6

Grav potential
symbol
eqn vector/scalar

Answer 6

ɸ, wk done by an eternal force in bringing a small test mass from infinity to that pt
ɸ = - GM/r

Question 7

What is the grav constant

Answer 7

G = 6.67 X 10^-11 N m^2 kg^-2

Question 8

2 ‘graphical’ eqn / r/s

Answer 8

F= - du/dr
g= - dɸ/dr

Question 9

Kepler’s 3rd Law and how to derive it

Answer 9

MUST SAY: the grav force on the planet provides the centripetal force

Sq of the period of revolution of the planets are directly prop to the cubes of their mean distances from the sun

By N2L, |Fg| = |Fc|
GMm/r^2 = mrω^2 
GM = r^3 ω^2
ω^2 = GM/r^3 = (2π/T)^2 = 4π^2/T^2
T^2 = 4πr^3/Gm --> T^2 ∝ r^3

Question 10

KE of orbiting satellite
GPE of orbiting satellite
Total energy of orbiting satellite

Answer 10

MUST SAY: the grav force on the planet provides the centripetal force
By N2L, |Fg| = |Fc|
GMm/r^2 = mv^2/r
1/2 mv^2 = GMm/2r

u= - GMm/r

Total energy = GMm/2r + (- GMm/r)
= -GMm/2r

Question 11

Min Escape Vel
Defn
layman ish meaning
derivation

Answer 11

Min speed required for obj to justttt escape from grav influence, any obj with a total energy of 0 will be able to just reach infinity and stop there

Basically when KE=U
If KE=GPE, orbiting obj/satellite will have just enough energy to escape from the g. influence of a massive body

-GMm/r + 1/2 mv^2 = 0
v^2 = 2GM/r
Vesc = √(2GM/r)

Question 12

Is Vesc dependent on mass of obj?

Answer 12

no. cuz Vesc = √(2GM/r)

Question 13

Characteristics of geostationary orbits/satellites

Answer 13

Fixed position in the sky

• share same axis of rotation as Earth
• Rotate from West to East
• Orbital period = 24h
• Only have 1 possible radius of orbit

Question 14

Apparent weight on equator n north pole

Answer 14

N is smallest at equator as
Fg-N = Fc
N = Fg - Fc
N  = Fg - mrω^2
Apparent weight is less than true weight (Fg)

At poles
Fg-N = Fc = mrω^2
Since r=0,
Fg - N = 0
Fg = N

When N=0,
Fg = Fc –> apparent weightlessness

Question 15

Resultant field strength at a pt due o more than 2 mass can be found…

Answer 15

Vector sum of the indiv grav field strengths

Question 16

For small distances above the surface of the earth, U ≈

Answer 16

U ≈ mgh

Question 17

Why is ɸ -ve?

Answer 17

ɸ at infinity is 0
Since F is an attractive force in nature, to bring a mass from infinity to a pt in the grave field, the direction of the external force is opp in direction of displacement of the mass.
This results in -ve wk done by the external force.
Hence, based on its defn, grav potential has a -ve value

Question 18

Adv and dis of geostationary satellites

Answer 18

• continues surveillance of region under it
• easy for the grd station to communicate as it is permanently in view. Grd-based antennas can remain fixed in 1 direction
• Due to high altitude, the satellite can transmit and receive signals over a large area

Disadv:

• as dist from earth’s surface is longer
• sig. loss of signal strengths
• poorer results in imaging satellites
• time lag in communications