A body released from a height h takes time to reach the earth's surface. The time taken by the same body releases from the same height to reach the moon's surface is
Answer
633.9k+ views
Hint: Objects that are said to be experiencing free fall do not encounter a major air resistance force; they fall under the sole influence of gravity. Acceleration due to gravity is called the acceleration of free-falling objects, since objects are pulled towards the center of the earth. Acceleration due to gravity on the Earth's surface is constant
Formula used:
$t=\sqrt{\dfrac{2h}{g}}$
Complete solution Step-by-Step:
Since the Earth is much larger (has greater mass), the gravitational force field of the Earth is stronger than the Moon's. A higher rate of acceleration (g = \[9.8m/{{s}^{2}}\]) is caused by the stronger gravitational force field on Earth than the weaker field of gravitational force on the Moon (g moon = \[1.6m/{{s}^{2}}\]).
An object that is falling under the sole influence of gravity is a free-falling object. Any object that is only acted on by the force of gravity is said to be in a free fall state. Air resistance is not experienced by free-falling objects.
If body falls from height h then time of descent
$t=\sqrt{\dfrac{2h}{g}}$
$\begin{aligned} t &=\sqrt{\dfrac{2 h}{g}} \Rightarrow \dfrac{t_{\text {moon}}}{t_{\text {carth}}}=\sqrt{\dfrac{g_{\text {earth}}}{g_{\text {moon}}}} \\ &=\sqrt{6} \Rightarrow t_{\text {moon}}=\sqrt{6} t \end{aligned}$
$\begin{aligned} t &=\sqrt{\dfrac{2 h}{g}} \Rightarrow \dfrac{t_{\text {moon}}}{t_{\text {carth}}}=\sqrt{\dfrac{g_{\text {earth}}}{g_{\text {moon}}}} \\ &=\sqrt{6} \Rightarrow t_{\text {moon}}=\sqrt{6} t \end{aligned}$
$\therefore {{t}_{\text{moon}}}=\sqrt{6}t$
The time taken by the same body released from the same height to reach the moon's surface is $\sqrt{6}t$.
Note:
Free-fall acceleration is often seen by means of an ever-popular demonstration of strobe light. As such, regardless of their mass, all objects free fall at the same rate. Since the gravitational field on the surface of Earth causes any object placed there to accelerate, we often call this ratio the acceleration of gravity. The object falls freely toward the earth, whereas its speed does not change in magnitude along the orbit and prevents it from hitting the earth. For two-body movement in a gravitational field, there are several possible scenarios.
Formula used:
$t=\sqrt{\dfrac{2h}{g}}$
Complete solution Step-by-Step:
Since the Earth is much larger (has greater mass), the gravitational force field of the Earth is stronger than the Moon's. A higher rate of acceleration (g = \[9.8m/{{s}^{2}}\]) is caused by the stronger gravitational force field on Earth than the weaker field of gravitational force on the Moon (g moon = \[1.6m/{{s}^{2}}\]).
An object that is falling under the sole influence of gravity is a free-falling object. Any object that is only acted on by the force of gravity is said to be in a free fall state. Air resistance is not experienced by free-falling objects.
If body falls from height h then time of descent
$t=\sqrt{\dfrac{2h}{g}}$
$\begin{aligned} t &=\sqrt{\dfrac{2 h}{g}} \Rightarrow \dfrac{t_{\text {moon}}}{t_{\text {carth}}}=\sqrt{\dfrac{g_{\text {earth}}}{g_{\text {moon}}}} \\ &=\sqrt{6} \Rightarrow t_{\text {moon}}=\sqrt{6} t \end{aligned}$
$\begin{aligned} t &=\sqrt{\dfrac{2 h}{g}} \Rightarrow \dfrac{t_{\text {moon}}}{t_{\text {carth}}}=\sqrt{\dfrac{g_{\text {earth}}}{g_{\text {moon}}}} \\ &=\sqrt{6} \Rightarrow t_{\text {moon}}=\sqrt{6} t \end{aligned}$
$\therefore {{t}_{\text{moon}}}=\sqrt{6}t$
The time taken by the same body released from the same height to reach the moon's surface is $\sqrt{6}t$.
Note:
Free-fall acceleration is often seen by means of an ever-popular demonstration of strobe light. As such, regardless of their mass, all objects free fall at the same rate. Since the gravitational field on the surface of Earth causes any object placed there to accelerate, we often call this ratio the acceleration of gravity. The object falls freely toward the earth, whereas its speed does not change in magnitude along the orbit and prevents it from hitting the earth. For two-body movement in a gravitational field, there are several possible scenarios.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Simon Commission came to India in A 1927 B 1928 C 1929 class 11 social science CBSE

How are involuntary actions and reflex actions different class 11 biology CBSE

Earth rotates in which direction A East to west B West class 11 physics CBSE

Study of snakes is called A Herpetology B Ichthyology class 11 biology CBSE

What are polar and nonpolar solvents class 11 chemistry CBSE

