Even gravity wants me!

Classical Mechanics Level 4

A body is projected upwards with twice the escape velocity on Earth, $11.2$ km/s.

What is the speed of the body at infinity? Ignore the presence of other heavenly bodies. Answer in km/s. Round to one decimal point.

The answer is 19.4.

4 solutions

Rajdeep Dhingra
Oct 29, 2014

Answer Answer

could you please rewrite your solution???the picture is damage

safa m - 6 years, 5 months ago

Rama Devi
May 18, 2015

If v is the velocity of projected and v' is the velocity at infinity, then we have by energy conservation principle.

½ mv2 - GMm/R = 1/2mv'2 + 0

Here v = 2ve

Thus, (1/2) . 4ve2 - GM/R = 1/2v'2

=> 2ve2 - GM/R = 1/2v'2

Now ve = √2GM/R

=> 2ve2 - v2e/2 = 1/2v'2

or, v'2 = 3 ve2

or, v' = √3 ve = √3 × 11.2 km/s = 19.4 km/s.

Revanth Gumpu
Jan 10, 2015

Let a = Velocity at projection. Let b = Velocity at infinity.

We can apply conservation of energy here since no energy is lost in a system. (ma^2)/2 - (GMm)/r = (mb^2)/2 + 0

We are given that the velocity is twice the escape velocity so can say the escape velocity is c which then means a = 2c. Plugging into the equation, we also see that m is common to all terms so divide by m and the equation becomes;

4c^2 - GM/r = b^2/2

Using this and the fact that c = √(2GM/r), we will get the final equation of b = √3 c. Solving we get: √3 × 11.2 km/s = 19.4 km/s.

3*11.2=19.3?

You should've wrote $\sqrt{3}11.2=19.3$

Muhammad Arifur Rahman - 6 years, 4 months ago

Somesh Meena
Jan 13, 2015

When the velocity is 11.2(escape velocity) then v^2=u^2 + 2as
=> 0=(11.2)^2 + 2as
Now taking velocity = 2×11.2
v^2 = (22.4)^2 + 2as
putting 2as = -(11.2)^2
We get v = 19.40

Even gravity wants me!

The answer is 19.4.

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