Find Acceleration due to Gravity for m = 1705 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 = 1705 t, r = 60 ft i.e. 3.4023 x 10-07 m/s2 in a short span of time.
Detailed Steps to Calculate Acceleration due to Gravity for m = 1705 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:
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 = 1705.0 t , and r = 60.0 ft :
Given that,
G = 6.674 x 10-11 N m2/kg2
M = 1705.0 t
r = 60.0 ft
⇒ Convert the Mass value 1705.0 t to "Kilograms (kg)"
Mass in Kg = 1705.0 ÷ 0.001
Mass in kg = 1705000.0 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 = 3.4023 x 10-07 m/s2
∴ Acceleration due to Gravity (a) = 3.4023 x 10-07 m/s2
Acceleration due to Gravity for m = 1705 t, r = 60 ft Results in different Units
Values | Units |
---|---|
3.4023 x 10-07 | meter per second squared (m/s2) |
3.4023 x 10-05 | centimeter per second squared (cm/s2) |
1.1163 x 10-06 | feet per second squared (ft/s2) |
2.1141 x 10-10 | mile per second squared (mi/s2) |
3.4023 x 10-10 | kilometer per second squared (km/s2) |
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