Find Acceleration due to Gravity for m = 1117 t, r = 60 ft

Created By : Sunil Kumar Gandipadala

Reviewed By : Rajashekhar Valipishetty

Last Updated : Apr 10, 2023

Take the help of Acceleration due to Gravity Calculator to determine acceleration (a) for m = 1117 t, r = 60 ft i.e. 2.229 x 10-07 m/s2 in a short span of time.

 

Ex: 23, 456, 2e4 , 2e-11, 1.23, 3.076e-11

Mass of the object (M)
Radius of the object (R)

Find Acceleration due to Gravity for m = 1117 t, r = 60 ft

Detailed Steps to Calculate Acceleration due to Gravity for m = 1117 t, r = 60 ft

The Acceleration due to earth gravity is known as the acceleration due to gravity. It means when an object falls from a certain height towards the surface of the earth, its velocity changes.

The formula to calculate the acceleration due to gravity is given by:

a =
G x M / r2

Where,

a = Acceleration due to Gravity

G = Gravitational constant value (i.e. 6.674 x 10-11)

M = Mass of an object

r = Radius from planet centre


Step by Step Solution to find acceleration due to gravity of M = 1117.0 t , and r = 60.0 ft :

Given that,

G = 6.674 x 10-11 N m2/kg2

M = 1117.0 t

r = 60.0 ft

⇒ Convert the Mass value 1117.0 t to "Kilograms (kg)"

Mass in Kg = 1117.0 ÷ 0.001

Mass in kg = 1117000.0 kg

=> Convert radius value 60.0 ft to "meters (m)"

Radius in meters = 60.0 ÷ 3.281

Radius in meters (m) = 18.28799941 m

Substitute the value into the formula

a =
6.674 x 10-11 x 1117000.0 / (18.28799941)2

a = 2.229 x 10-07 m/s2

∴ Acceleration due to Gravity (a) = 2.229 x 10-07 m/s2

Acceleration due to Gravity for m = 1117 t, r = 60 ft Results in different Units
Values Units
2.229 x 10-07 meter per second squared (m/s2)
2.229 x 10-05 centimeter per second squared (cm/s2)
7.3129 x 10-07 feet per second squared (ft/s2)
1.385 x 10-10 mile per second squared (mi/s2)
2.229 x 10-10 kilometer per second squared (km/s2)

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