Question

There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2. What is the minimum speed v which will permit the projectile to reach station Beta?

Answer 1

The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]

What is gravitational potential energy?

If an object is lifted, work is done against gravitational force. The object gains energy.

Given are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.

The range of the projectile is given by R =  v²sin2θ / g

g = gravitational acceleration of Earth

If g = g(p) for planet , range  R =  v²sin2θ / g(p)..................(1)

The gravitational force of attraction = weight force

Gm² /d² = m g(p)

g(p) = Gm/d².........................(2)

For R = d/3, from equation (1), we have

d/3 =  v²sin2θ / g(p)

Plug the expression for g(p) , we get

v = √ [Gm/3dsin2θ ]

For velocity to be minimum, sin2θ =1

So, the minimum velocity will be

v = √ [Gm/3d]

Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]

Learn more about gravitational potential energy

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