State Newton’s universal law of gravitation.

Deduce that the gravitational field strength g at the surface of a spherical planet of uniform density is given by

\[g = \frac{{GM}}{{{R^2}}}\]

where M is the mass of the planet, R is its radius and G is the gravitational constant. You can assume that spherical objects of uniform density act as point masses.

The gravitational field strength at the surface of Mars g_M is related to the gravitational field strength at the surface of the Earth g_E by

g_M = 0.38 × g_E.

The radius of Mars R_M is related to the radius of the Earth R_E by

R_M = 0.53 × R_E.

Determine the mass of Mars M_M in terms of the mass of the Earth M_E.

(i) On the diagram below, draw lines to represent the gravitational field around the planet Mars.

(ii) An object falls freely in a straight line from point A to point B in time t. The speed of the object at A is u and the speed at B is v. A student suggests using the equation v=u+g_Mt to calculate v. Suggest two reasons why it is not appropriate to use this equation.

there is an attractive force;
between any two point/small masses;
proportional to the product of their masses;
and inversely proportional to the square of their separation;
Accept formula with all terms defined.

use of \(g = \frac{F}{m}\) and \(F = \frac{{GmM}}{{{R^2}}}\);
evidence of substitution/manipulation;
to get \(g = \frac{{GM}}{{{R^2}}}\)

\(\frac{{{g_M}}}{{{g_E}}} = \frac{{\frac{{{M_M}}}{{R_M^2}}}}{{\frac{{{M_E}}}{{R_E^2}}}} \Rightarrow \frac{{{M_M}}}{{{M_E}}} = \frac{{{g_M}}}{{{g_E}}} \times {\left[ {\frac{{{R_M}}}{{{R_E}}}} \right]^2}\);

\({M_{\rm{M}}}\left( { = 0.38 \times {{0.53}^2}{M_{\rm{E}}}} \right) = 0.11{M_{\rm{E}}}\);

(i) radial field with arrows pointing inwards;

(ii) field between A and B is not equal to field at surface;
acceleration is not constant between these two points;