Question

Space explorers discover an 8.7 × 1017 kg asteroid that happens to have a positive charge of 4400 C. They would like to place their 3.3 × 105 kg spaceship in orbit around the asteroid. Interestingly, the solar wind has given their spaceship a charge of -1.2 C.
What speed must their spaceship have to achieve a 8300-km-diameter circular orbit?

Answer 1

We have that the speed of the spaceship must be

$V=2.47m/s$

From the Question we are told that

Mass $8.7*10^{17} kg$

Positive charge of 4400 C

Spaceship mass m_s=3.3* 10^5 kg

Solar wind has given their spaceship a charge of -1.2 C.

Diameter d=8300-km-diameter

Generally the equation for the force of attraction   is mathematically given as

$v^2=\frac{Gm}{r}+\frac{kQq}{rm}\\\\v^2=\frac{6.67*10^{-11}*8.7*10^{17}}{9.5*10^5}}+\frac{9*10^9*4400*-1.2}{9.5*10^5*3.3*10^{15}}$

$V=2.47m/s$

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Answer 2

The speed will be "6.98 m/s". A complete solution is provided below.

The given values are:

Mass of asteroid,

• M = $8.7\times 10^{17}$ kg

Mass of spaceship,

• m = $3.3\times 10^{5}$ kg

Charge,

• $q_1=4400 \ C$
• $q_2=1.2 \ C$

As we know,

→ $F_G +F_E = \frac{mv^2}{R}$

or,

→ $\frac{GMm}{R_2} +\frac{kq_1 q_2}{R_2}=\frac{mv^2}{R}$

By putting the given values, we get

→ $\frac{6.67\times 10^{-11}\times 8.7\times 10^{17}\times 3.3\times 10^5}{\frac{8.3\times 10^6}{2} } +\frac{9\times 10^9\times 4400\times 1.2}{\frac{8.3\times 10^6}{2} } = 3.3\times 10^5\times v^2$

→                                                                   $v=6.98 \ m/s$                                                                                                                                                      

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