• A gravitational field is an example of a field of force
• Gravitational field strength is the gravitational force per unit mass

Describing a Gravitational Field

• For an isolated point mass, the gravitational field is spherical in shape with the mass at the center
• The gravitational field is described by the field lines.
• A field line is the path followed by a free unit mass in that gravitational field
• A higher density of field lines = a region of stronger field

Newton’s Law of Gravitation

• Gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation
  • F = ^GMm/_r²
  • G is the gravitational field constant and it is equal to 6.67 × 10^-11 Nm²kg^–²
• The gravitational force between two masses is independent of the medium separating the mass and is always an attractive force

Examples

1. The orbit of the Earth, mass 6.0 × 10²⁴ kg, may be assumed to be a circle of radius 1.5 × 10¹¹ m with the Sun at its center, illustrated below. The time taken for one orbit is 3.2 × 10⁷ s
   1. Calculate the magnitude of the centripetal force acting on the earth
      • Calculate the angular velocity of the earth:
        • ω = ^2π/_T
        • = ^2π/_{3.2 × 10⁷}
        • ω = 1.96 × 10^-7
      • Use the centripetal force equation:
        • F = mω²r
        • F = 6.0 × 10²⁴ × (1.96 × 10^-7)² × 1.5 × 10¹¹
        • ∴ F = 3.46 × 10²² N
   2. Determine the mass of the Sun
      • The centripetal force is provided by the gravitational force of the sun. Using Newton’s inverse law:
        • 3.46 × 10²² = ^GMm/_r²
      • Substitute values into the expression:
        • 3.46 × 10²² = ^{6.67 × 10-11 × M × 6.0 × 1024}/_{(1.5 × 10¹¹)²}
        • ∴ M = 1.95 × 10³⁰

Gravitational Field Strength

• The gravitational field strength at a point is the gravitational force exerted per unit mass
• By equating W = mg and Newton’s Law of Gravitation:
  • mg = ^GMm/_r²
  • ∴ g = ^GM/_r²

Gravitational Potential

• The gravitational potential at a point is work done per unit mass in bringing a mass from infinity to the point
  • Φ = –^GM/_r
• The negative sign is because:
  • Gravitational force is always attractive
  • Gravitational potential reduces to zero at infinity
  • Gravitational potential decrease in the direction of the field
    
• On Earth’s surface, we can use the equation g.p.e = mgh, however, this is not true for masses far from Earth’s surface because we assume g is constant
• Gravitational Potential Energy of a mass m at a point in the gravitational field of another mass M, is the work done in bringing that mass m from infinity to that point
  • U = mΦ = –^GM/_rm
• The gravitational potential energy difference between two points is the work done in moving a mass from one point to another
  • ΔU = mΦ_final – mΦ_initial

Centripetal Acceleration

• For an orbiting satellite, gravity provides centripetal force which keeps it in orbit
• Therefore,
  • ^GMm/_r² = ^mv²/_r
  • v² = ^GM/_r
• Velocity is independent of the mass of the satellite

Geostationary Orbits

• A geostationary orbit:
  • is an equatorial orbit
  • has a period of 24 hours; same angular speed as the Earth
  • moves from West to East; same direction or rotation as the Earth
• Geostationary Satellite is one which is always above a certain point on the Earth
• For a geostationary orbit, T = 24 hrs and the orbital radius is a fixed value from the center of the Earth
• However, the mass of the satellite is not fixed, hence the k.e., g.p.e, and centripetal force are not fixed values
• A geostationary satellite is launched from the equator in the direction of rotation of the Earth (West to East) so that the axis of rotation of the satellite and Earth coincide

Examples

1. The Earth may be considered to be a sphere of radius 6.4 × 10⁶ m with mass of 6.0 × 10²⁴ kg concentrated at its center. A satellite of mass 650 kg is to be launched from the equator and put into geostationary orbit
   1. Show that the radius of the geostationary orbit is 4.2 × 10⁷ m
      • Centripetal force provided by gravity:
        • ^GMm/_r² = ^mv²/_r
        • GM = v²r
      • Using angular velocity, substitute v = ωr
        • GM = ω²r² × r
      • Substituting ω = 2π/T:
        • GM = ^4π²/_T² × r³
      • The time period is always 24 hours, so in seconds:
        • 24 hours = 24 × 60 × 60 = 86400 s
      • Rearranging and substituting values:
        
   2. Determine the increase in gravitational potential energy of the satellite during its launch from the Earth’s surface to the geostationary orbit.
      • Using the following expression:
        
      • Substitute values:
        

Escape Velocity of a Satellite

• By conservation of energy, initial k.e. + initial g.p.e. = final = 0
  • ½mv² – ^GMm/_r = 0
  • Thus escape velocity = √(^2GM/_r)
• Escape velocity is the speed a satellite needs to get into orbit, however it is not used as it a huge value and satellites have engines to provide thrust to reach the height of the orbit

Weightless

• An astronaut is in a satellite orbiting the Earth, reports that he is ‘weightless’, despite being in the Earth’s gravitational field.
• This sensation is because:
  • Gravitational force provides the centripetal force: the gravitational force is equal to the centripetal force
    • ^GMm/_r² = ^mv²/_r
  • The sensation of weight (reaction force) is the difference between F_G and F_C which is zero
  • Therefore,astronauts feel weightless