Answer:
a) r = 4.22 10⁷ m, b) v = 3.07 10³ m / s and c) a = 0.224 m / s²
Explanation:
a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force
F = m a
a = v² / r
F = G m M / r²
G m M / r² = m v² / r
G M / r = v²
The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships
v = d / t
As the orbit is circular the distance is the length of the circle in 24 h time
d = 2π r
t = 24 h (3600 s / 1 h) = 86400 s
Let's replace
G M / r = (2π r / t)²
G M = 4 π² r³ / t²
r = ∛(G M t² / (4π²)
r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)
r = 4.22 10⁷ m
b) the speed module is
v = √G M / r
v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷
v = 3.07 10³ m / s
c) the acceleration is
a = G M / r²
a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²
a = 0.224 m / s²
Answer:
The answer is D (It reduced his average
force.)
<span>Electromagnetic, Strong Nuclear, Weak Nuclear, and Gravity</span>
