A one tonne moon obiter is to approach the moon and orbit through 270 degrees at an altitude of 100km. During this period its rockets are to be turned off. What is the required velocity of the obiter at the entry point to prevent the vehicle from crashing into the moon?
centripetal acceleation=gravitational acceleation
v^2/r= GMassmoon/r^2
where r is the radius of the moon+100km
solve for v
Thanks a lot
To determine the required velocity of the orbiter at the entry point to prevent it from crashing into the moon, we need to consider the gravitational force between the orbiter and the moon. The gravitational force provides the necessary centripetal force to maintain an orbit.
First, let's consider the gravitational force equation:
F = G * ((m1 * m2) / r^2)
Where:
F is the gravitational force
G is the gravitational constant (approximately 6.67430 × 10^-11 N(m/kg)^2)
m1 is the mass of the moon (approximately 7.349 × 10^22 kg)
m2 is the mass of the orbiter (one tonne = 1000 kg)
r is the distance between the moon's center and the obiter's center (altitude + radius of the moon)
Given:
Mass of the orbiter (m2) = 1000 kg
Altitude (100 km) = 100,000 m
Radius of the moon = 1,737,100 m (approximately)
Now, to find the required velocity, we need to calculate the orbital speed at an altitude of 100 km.
Orbital speed (V) can be calculated using the formula:
V = sqrt(G * (m1 + m2) / r)
Let's substitute the known values into the formula:
V = sqrt(6.67430 × 10^-11 * (7.349 × 10^22 + 1000) / (100,000 + 1,737,100))
Calculating this will give us the required velocity (V) at the entry point to prevent the orbiter from crashing into the moon.
Please note that the given values are approximate and the actual calculations may vary slightly depending on the precise values used.