Answer:
u = 12962.11 m/s
Explanation:
Given that,
The radius of mercury, [tex]r=2440\ km=2440\times 10^3\ m[/tex]
Mass of Mercury, [tex]M=3\times 10^{24}\ kg[/tex]
Final speed of the object, v = 2000 m/s
Let u is its initial speed when the object is far from Mercury. It can be calculated by applying the conservation of energy as :
Initial kinetic energy + gravitational potential energy = final kinetic energy
[tex]\dfrac{1}{2}mu^2+(-\dfrac{GmM}{r})=\dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}u^2+(-\dfrac{GM}{r})=\dfrac{1}{2}v^2[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+\dfrac{GM}{r}[/tex]
[tex]u^2=2\times (\dfrac{1}{2}v^2+\dfrac{GM}{r})[/tex]
[tex]u^2=2\times (\dfrac{1}{2}(2000)^2+\dfrac{6.67\times 10^{-11}\times 3\times 10^{24}}{2440\times 10^3})[/tex]
u = 12962.11 m/s
So, the initial speed of the object is 12962.11 kg. Hence, this is the required solution.
