Gravitational fields

Question 1

Define

gravitational field

Answer 1

A region of space where a mass experiences a gravitational force

A mass falling in a gravitational field will fall parallel to the field.

Question 2

Define

gravitational field strength

Answer 2

Force per unit mass

g = F/m (N kg^-1)

Question 3

State

Newton’s law of gravitation

Answer 3

Any 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 their separation.

F = -GMm/r²

For a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre.

The negative sign of the force indicates the force is attractive.

Question 4

Derive the equation for the gravitational field strength of a point mass

Answer 4

g = F/m and F = -GMm/r²

Therefore, g = -GMm/r²m (cancel m)

g = -GM/r²

Question 5

Define

gravitational potential

Answer 5

The gravitational potential at a point is the work done in moving unit mass from infinity to the point.

φ = W/m = -GM/r

φ is always negative as at infinity φ = 0 (which is the maximum), and forces between masses are attractive (the minus sign can be proven using calculus).

Δφ is negative moving towards a mass and positive moving away from a mass.

Question 6

Derive the formula for gravitational potential energy

Answer 6

work done = force x distance

W = -GMm/r² x r

W = -GMm/r

OR

g.p.e. = mgh

W = m x GM/r² x r

W = -GMm/r

Question 7

Derive a formula for the change in gravitational potential energy of a mass m from the surface of a large mass M to a point in space

Answer 7

At surface: E = -GMm/R

In space: E = -GMm/r

ΔE = -GMm/r - -GMm/R

ΔE = GMm(1/R - 1/r)

also equivalent to:

ΔE = GMm(1/R - 1/(h+R))

Question 8

Derive a formula for the square of the period of Earth’s orbit, and hence prove Kepler’s 3^rd law

Answer 8

gravitational force = centripetal force

GM_SM_E/R² = M_ERω²

GM_S/R² = R (2π/T)²

T² = (R x 4π² x R²) / (GM_S)

T² = (4π²/GM_S) R³

Therefore, T² ∝ R³

Question 9

Derive a formula for the kinetic energy of a satellite in orbit

Answer 9

centripetal force = gravitational force

mv²/r = GMm/r²

v² = GM/r

Substituting into E_K = 1/2 mv²

E_K = 1/2 GMm/r

Question 10

Derive a formula for the total energy of a satellite in orbit

Answer 10

E_T = E_K + E_P

E_T = 1/2 GMm/r + -GMm/r

E_T = -1/2 GMm/r

Question 11

Derive the formula v = √(2gr) for the escape velocity of a satellite from gravity to infinity

Answer 11

kinetic energy = change in g.p.e.

1/2 mv² = GMm/r

v = √(2GM/r)

v = √(2GM/r² x r)

v = √(2gr)

Question 12

Define

1. geostationary orbit
2. circular polar orbit

and state why each is used

Answer 12

1. The satellite orbits Earth at the same speed of Earth’s rotation, enabling it to stay above the same spot on Earth (useful for monitoring weather, communications and surveillance)
2. The satellite passes above or nearly above both poles of Earth (often used for earth-mapping, earth observation, capturing the earth as time passes from one point, reconnaissance satellites, as well as for some weather satellites)