Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad
1 answer:
Answer:
Force of gravity, F = 0.74 N
Mass of an object, m = 1.2 kg
Distance above earth's surface,
Mass of Earth,
Radius of Earth,
We need to find the force of gravity above the surface of Earth. It is given by :
R = r + d
R = 25370000 m
F = 0.74 N
So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.
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Answer:
27.5 days
0.92 month
Explanation:
= radius of the orbit of moon around the earth =
= Mass of earth =
= Time period of moon's motion
According to Kepler's third law, Time period is related to radius of orbit as
inserting the values, we get
we know that
1 day = 24 hours = 24 x 3600 sec = 86400 s
1 month = 30 days
Answer:
v = 1176.23 m/s
y = 741192.997 m = 741.19 km
Explanation:
Given
M₀ = 9 Kg (Initial mass)
me = 0.225 Kg/s (Rate of fuel consumption)
ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)
v = ? if t = 20 s
y = ?
We use the equation
v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)
⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t
then we have
v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)
v = 1176.23 m/s
then we apply the formula
y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt
⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt
⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)
For t = 20 s we have
y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)
⇒ y = 741192.997 m = 741.19 km
The graphs are shown in the pics.
