The mean radius of the Earth is R=6.38x10^6m. At what distance above the Earth's surface will the acceleration of gravity be 4.905m/s^2?
F_g=(Gmme)/r^2
4.905=6.67x10^-11(5.98x10^24)/r^2
The mass cancel out so that is how I got the above answer.
r=9017663m
Is this the correct answer or would I have to take 9017663m minus the radius of the earth to figure out the distance above the earth's surface. When i did this I got 2637663m Is this the correct answer or is 9017663m?
To find the distance above the Earth's surface at which the acceleration of gravity is 4.905 m/s^2, we can use the formula for gravitational acceleration:
F_g = (G * m_e * m) / r^2
Where F_g is the acceleration of gravity, G is the gravitational constant (6.67 x 10^-11 N m^2/kg^2), m_e is the mass of the Earth (5.98 x 10^24 kg), m is the mass of the object, and r is the distance from the center of the Earth.
We know that F_g is 4.905 m/s^2 and m_e cancels out, so we're left with:
4.905 = (G * m) / r^2
Rearranging the equation, we get:
r^2 = (G * m) / 4.905
Substituting the values of G and rearranging, we have:
r^2 = (6.67 x 10^-11 * m) / 4.905
Next, substitute the given value of the Earth's mean radius, R = 6.38 x 10^6 m, and solve for m:
(6.38 x 10^6)^2 = (6.67 x 10^-11 * m) / 4.905
Simplifying the equation, we find:
m = (4.905 * (6.38 x 10^6)^2) / 6.67 x 10^-11
m ≈ 2.0275 x 10^11 kg
Now that we have the mass of the object, we can substitute it back into the equation for r:
r^2 ≈ (6.67 x 10^-11 * 2.0275 x 10^11) / 4.905
r ≈ √(13.57325 x 10^3) ≈ 116.4639 x 10^3 m
So, the distance above the Earth's surface at which the acceleration of gravity is 4.905 m/s^2 is approximately 116,463.9 km.
To find the distance above the Earth's surface, subtract the radius of the Earth (6.38 x 10^6 m) from this distance:
116,463.9 km - 6.38 x 10^6 m = 110,083.9 km
Therefore, the correct answer is approximately 110,083.9 km above the Earth's surface.