Question

Starting from the formula for gravitational potential energy: U_G = Gm_1 m_2/r, derive the universal law of gravitation: F vector_G = Gm_1 M_2/r^2 f. Starting from the definition of the resultant force being the time derivative of translational momentum, show that the resultant force is only equal to ma vector if dm/dt = 0.

Answer 1

a)

gradient of the potential anergy gives the force,

Gradient of the potential anergy gives the force, F = -dU_G/dx Putting potential energy in the above term will get, F = -d/dr(Gm_1m_2/r) Finding the differentiation, F = Gm_1m_2/r^2 b) Here force is Given by, F = dp/dt where p is the momentum, F = d(m nu)/dt Using the product rule, F = m d(nu)/dt + nud(m)/dt Form above it is very much clear that F = ma is the valid expression if and only if we have dm/dt = 0. F = ma[because dm/dt = 0]

Putting potential energy in the above term will get,

$F=-\frac{d}{dr}\left ( \frac{Gm_1m_2}{r} \right )$

Finding the differentiation,

$F=\frac{Gm_1m_2}{r^2}$

b)

Here force is given by,

$F=\frac{dp}{dt}$

where p is the momentum,

$F=\frac{d(mv)}{dt}$

Using the product rule,

$F=m\frac{d(v)}{dt}+v\frac{d(m)}{dt}$

Form above it is very much clear that F=ma is the valid expression if and only if we have dm/dt =0.

$F=ma \left [ \because \frac{dm}{dt}=0 \right ]$