The gravitational force between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of
, while the second "object" is the Earth, with mass
. The distance of the object from the Earth's center is
; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
The correct answer is letter D. candela. The unit for measuring the rate at which light energy is radiated from a source is the candela. L<span>umen is the unit for measuring the total amount of visible light emitted by a source. Lux is lumen per square meter. </span>
Answer:
Explanation:
On the Moon :----
1500 x 1.6 = 2400 m /s is initial velocity of bullet .
g = 1.6 m /s²
v = u - gt
0 = 2400 - 1.6 t
t = 1500 s
This is time of ascent
Time of decent will also be the same
Total time of flight = 2 x 1500 = 3000 s
On the Earth : ---
v = u - a₁ t
0 = u - a₁ x 18
u = 18a₁
v² = u² - 2 x a₁ x 2743.2
0 = (18a₁ )² - 2 x a₁ x 2743.2
a₁ = 16.93
For downward return
s = ut + 1/2 a₂ x t²
2743.2 = 0 + .5 x a₂ x 31²
a₂ = 5.7 m /s²
If d be the deceleration produced by air
g + d = 16.93 ( during upward journey )
g - d = 5.7
g = (16.93 + 5.7) / 2
= 11.315 m / s
d = 5.6 m /s²
So air is creating a deceleration of 5.6 m /s².
