Question

Deimos completes one (circular) orbit of Mars in 1.26 days. The distance from Mars to Deimos is 2.35×107m. What is the centripetal acceleration of Deimos?

Answer 1

Answer:

The centripetal acceleration of Deimos is $0.077m/s^{2}$.

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$   (1)

Where v is the velocity of Deimos and r is the orbital distance.        

Notice that is necessary to determine the velocity first.

The speed of the Deimos can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$  (2)

Then, replacing Newton's second law in equation 2 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$  (3)

However, a is the centripetal acceleration since Deimos almost describes a circular motion around Mars:

$a = \frac{v^{2}}{r}$  (4)

Replacing equation 4 in equation 3 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$  (5)

Where v is the orbital speed, G is the gravitational constant, M is the mass of Mars, and r is the orbital radius.    

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(6.39x10^{23}kg)}{2.35x10^{7}m}}$

$v = 1346m/s$

Finally, equation 4 can be used:

$a = \frac{(1346m/s)^{2}}{2.35x10^{7}m}$

$a = 0.077m/s^{2}$

Hence, the centripetal acceleration of Deimos is  $0.077m/s^{2}$.