Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
46 m and 159 m
Explanation:
first find initial x and y component of velocity and take final y component of velocity as zero at maximum point and follow steps as i have solved
Answer:
1.5 hr
16.7
Explanation:
Zero apparent weight means there's no normal force.
Sum the forces in the centripetal direction.
∑F = ma
mg = mv²/r
v = √(gr)
v = √(7.4×10⁶ m × 10 m/s²)
v = 8602 m/s
The circumference of the equator is:
C = 2πr
C = 2π (7.4×10⁶ m)
C = 4.65×10⁷ m
So the period is:
T = C / v
T = (4.65×10⁷ m) / (8602 m/s)
T = 5405 s
T = 1.5 hr
The initial speed is:
v = C / T
v = (4.65×10⁷ m) / (25 h × 3600 s/h)
v = 517 m/s
The speed increases by a factor of:
8602 m/s / 517 m/s
16.7