Question

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 5.00�1015kg and a radius of 8.80km .A.What is the speed of a satellite orbiting 5.20km above the surface?B.What is the escape speed from the asteroid?

Answer 1

Answer:

a). What is the speed of a satellite orbiting 5.20km above the surface?

The speed of the satellite is $4.88m/s$.

b). What is the escape speed from the asteroid?

The escape speed from the asteroid is $6.90m/s$.

Explanation:

a). What is the speed of a satellite orbiting 5.20km above the surface?

The speed of the satellite can be found through the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$  (1)

Then, replacing Newton's second law in equation 1 it is gotten:

$m.a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a is the centripetal acceleration since the satellite describes a circular motion around the asteroid:

$a = \frac{v^{2}}{r}$  (3)

Replacing equation 3 in equation 2 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$  (4)

Where v is the orbital speed, G is the gravitational constant, M is the mass of the asteroid, and r is the orbital radius.

Notice that the orbital radius will be given by the sum of the radius of the asteroid and the height of the satellite above the surface:

$r = 8.80km+5.20km$

$r = 14km$

But it is necessary to express r in units of meters before it can be used in equation 4.

$r = 14km \cdot \frac{1000m}{1km}$ ⇒ $14000m$

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(5.00x10^{15}kg)}{14000m}}$

$v = 4.88m/s$

Hence, the speed of the satellite is $4.88m/s$

b). What is the escape speed from the asteroid?

The escape speed is defined as:

$v_{e} = \sqrt{\frac{2GM}{r}}$ (5)

Where G is the gravitational constant, M is the asteroid radius and r is the orbital radius.

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.00x10^{15}kg)}{14000m}}$

$v_{e} = 6.90m/s$

Hence, the escape speed from the asteroid is $6.90m/s$.

Answer 2

Answer:

a).$v_{1}=13.49 \frac{m}{s}$

b).$v_{2}=17.54\frac{m}{s}$

Explanation:

Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in

$\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}$

$v^2=\frac{2*G*M*m}{a_{o}}$

$G=6.673x10^{-11}\frac{m^3}{kg*s^2}$

$M=1.20x10^{16}kg$

a).

$a_{o}=8.8km*\frac{1000m}{1km}=8800m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}$

$v_{1}=13.49 \frac{m}{s}$

b).

$a_{f}=5.2km*\frac{1000m}{1km}=5200m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}$

$v_{2}=17.54\frac{m}{s}$