Question

(A) How do you explain that an object is in uniform circular motion?
(B) Calculate the acceleration of the moon towards earth’s centre.

Answer 1

Hint:Give the information about the uniform circular motion of an object. Use the formula for centripetal force and expression for Newton’s gravitational law of attraction between the two objects. Using these two formulae, calculate the centripetal acceleration of the moon towards the centre of the Earth.

Formulae used:
The centripetal force \[{F_C}\] on an object is
\[{F_C} = \dfrac{{m{v^2}}}{R}\] …… (1)
Here, \[m\] is the mass of the object, \[v\] is the velocity of the object and \[R\] is the radius of the circular path.
The expression for Newton’s law of gravitational attraction is
\[F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}\] …… (2)
Here, \[F\] is force of attraction between two object, \[G\] is universal gravitational constant, \[{m_1}\] and \[{m_2}\] are the masses of the two objects and \[r\] is distance between the centres of the two objects.

Complete step by step answer:
(A) An object is said to be in uniform circular motion if it is moving with a constant speed on the circumference of a circular track.Although the speed of the object in the uniform circular motion is constant, the velocity of the object is not constant.Hence, the acceleration of the object changes continuously due to continuous change in direction of motion of the object.

For an object in the uniform circular motion, the centripetal acceleration of the object is always directed towards the centre of the circular path.
This centripetal acceleration \[{a_C}\] is given by
\[{a_C} = \dfrac{{{v^2}}}{R}\]
Here, \[v\] is the velocity of the object and \[R\] is the radius of the circular path.

(B) We have now asked to calculate the acceleration of the moon acting towards the centre of the Earth.Rewrite equation (1) for the centripetal force \[{F_C}\] acting on the moon while revolving around the Earth.
\[{F_C} = \dfrac{{m{v^2}}}{R}\]
Here, \[m\] is the mass of the moon, \[v\] is velocity of the moon around the Earth and \[R\] is the distance between the centres of the moon and the Earth.

Rewrite equation (2) for the gravitational force \[F\] of attraction between the moon and the Earth.
\[F = \dfrac{{GMm}}{{{R^2}}}\]
Here, \[M\] is the mass of the Earth.
For the moon revolving around the Earth, the centripetal force acting on the moon is equal to the gravitational force of attraction between the Earth and the moon.
\[{F_C} = F\]
Substitute \[\dfrac{{m{v^2}}}{R}\] for \[{F_C}\] and \[\dfrac{{GMm}}{{{R^2}}}\] for \[F\] in the above equation.
\[\dfrac{{m{v^2}}}{R} = \dfrac{{GMm}}{{{R^2}}}\]
\[ \Rightarrow \dfrac{{{v^2}}}{R} = \dfrac{{GM}}{{{R^2}}}\]

Substitute \[{a_C}\] for \[\dfrac{{{v^2}}}{R}\] in the above equation.
\[ \Rightarrow {a_C} = \dfrac{{GM}}{{{R^2}}}\]
Substitute \[6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}}\] for \[G\], \[5.97 \times {10^{24}}\,{\text{kg}}\] for \[M\] and \[384.4 \times {10^6}\,{\text{m}}\] for \[R\] in the above equation.
\[ \Rightarrow {a_C} = \dfrac{{\left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}}} \right)\left( {5.97 \times {{10}^{24}}\,{\text{kg}}} \right)}}{{{{\left( {384.4 \times {{10}^6}\,{\text{m}}} \right)}^2}}}\]
\[ \therefore {a_C} = 2.69 \times {10^{ - 3}}\,{\text{m/}}{{\text{s}}^2}\]

Hence, the acceleration of the moon towards the centre of the Earth is \[2.69 \times {10^{ - 3}}\,{\text{m/}}{{\text{s}}^2}\].

Note:The students may think that we have not considered the tangential acceleration of the moon while calculating its acceleration around the Earth. But the motion of the moon around the Earth is uniform circular motion and in uniform circular motion, the tangential acceleration is zero. Also we have asked to calculate the acceleration of the moon towards the centre of the Earth which is the direction of the centripetal acceleration.