Gravitational Field

Question 1

Newton’s law of Gravitation

Answer 1

the mutual force of attraction between any two point masses is directly proportional to the product of their masses and inversely proportional to the square of the separation between their centres

F = G m₁ m₂ / r²

F magnitude of gravitational force [N]
G gravitational constant 6.67 × 10-¹¹ N m² kg⁻¹
m₁ one of the point masses [kg]
m₂ other point mass [kg]
r centre to centre distance between the two point masses [m]

Question 2

Gravitational field

Answer 2

Region of space surrounding a body in which another body placed in it experiences a gravitational force due to its mass

Question 3

Gravitational field strength g at a particular point in the gravitational field

Answer 3

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

g = GM/ r²

g magnitude of gravitational field strength at point [N kg⁻¹]
G gravitational constant 6.67 × 10⁻¹¹ N m² kg⁻²
M source mass that generates the gravitational field [kg]
r distance from centre of source mass M to point X [m]

Question 4

Gravitational potential energy U of a point mass placed at a point in the gravitational field

Answer 4

work done in bringing the point mass from infinity to that point

Question 5

Gravitational potential energy U of the M-m system [J]

Answer 5

U = - GMm/r

G gravitational constant 6.67 × 10⁻¹¹ N m² kg⁻²
M one of the point mass [kg]
m other point mass [kg]
r centre-to-cetre distance between the two point masses [m]

Question 6

The escape speed/ velocity for an object at the Earth’s surface

Answer 6

minimum speed it must have at the start to escape from the gravitational influence of the Earth

v = √( 2GMe/Re)

Question 7

Gravitational potential φ at a point in the gravitational field

Answer 7

work done per unit mass in bringing a small test mass from infinity to that point

φ = -GM/r

φ gravitational potential at point X [ J kg⁻¹]
G gravitational constant 6.67 × 10⁻¹¹ N m² kg⁻²
M source mass that generates the gravitational field [kg]
r distance from centre of point mass M to point X [m]

Question 8

relation of F to U

Answer 8

F - δU/ δr

Question 9

relation of g to φ

Answer 9

g = - δφ/δr

Question 10

orbiting speed of satellite

Answer 10

Fg provides Fc

v = √ ( GM /r)

Question 11

energy of satellite in orbit

Answer 11

Ek = 1/2 m v²
= GMm/2r

Ep = - GMm/r

Et = - GMm/2r

Et = - Ek
= 1/2 Ep

Question 12

Kepler’s third law

Answer 12

Fg provides Fc

GMm/r² = mrω² , ω = 2π/T

T² = 4π²r³/GM

T² ∝ r³

Question 13

Geostationary satellite

Answer 13

rotates around planet such that it is always positioned above same point on the planet’s surface

Question 14

Conditions for satellite moving in geostationary orbit for Earth

Answer 14

1. placed above equator - same axis of rotation
2. move from west to east - same direction as rotation of Earth about its own axis
3. satellite orbital period = 24h - same as Earth’s rotational period about it’s own axis

Question 15

Advantages and disadvantages of geostationary satellite

Answer 15

Advantages

1. same relative position to Earth’s surface, allows continuous and uninterrupted signal transmission- telecommunications
2. positioned at high altitudes, scan a large section of Earth’s surface - weather imaging

Disadvantages

1. high altitudes ; poor resolution, delay in reception hence lag, expensive to launch
2. weak signal received by transmitting stations of countries with latitudes above 60° due larger amount of atmosphere