Question

An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 kg. An experiment is planned where a projectile needs to be launched straight up from the surface.

What must be the minimum initial speed of the projectile so it will reach a height of 2.55 × 10^6 m above this satellite's surface? (G = 6.67 × 10^-11 N · m^2/kg^2)

Answer 1

The minimum initial speed of the projectile to reach a height of 2.55 × 10^6 m above the satellite's surface can be calculated using the concept of gravitational potential energy.

Step-by-step explanation:

To calculate the minimum initial speed of the projectile to reach a height of 2.55 × 10^6 m above the satellite's surface, we can use the concept of gravitational potential energy. The gravitational potential energy at the surface of the satellite can be calculated using the formula:

PE = -GMm/r, where G is the gravitational constant, M is the mass of the satellite, m is the mass of the projectile, and r is the radius of the satellite.

At the highest point of the projectile's trajectory, all of its initial kinetic energy will have been converted to potential energy. So, we can equate the initial kinetic energy to the change in potential energy:

KE = PE

1/2 mv^2 = -GMm/r + GMm/(r + h), where v is the initial speed of the projectile and h is the vertical height above the surface.

By substituting the given values, we can solve for v:

v = sqrt((2GM/r) + (2GM/(r + h))).

Answer 2

2.87 km/s

Step-by-step explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

`g=(GM)/(R^(2))`

`g=(6.67* 10^(-11)* 7.35* 10^(22))/(1.74^(2)* 10^(12))`

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

`v^(2)=u^(2)-2gh`

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.