# Hi please explain it step by steps. Thank you. A particle is projected vertically upwards from the surface of the Earth with an initial velocity of U. On neglecting the resistance of the...

Hi please explain it step by steps.

Thank you.

A particle is projected vertically upwards from the surface of the Earth with an initial velocity of U. On neglecting the resistance of the atmosphere, the velocity, *v *, of the particle satisfies the differential equation `V(dv)/dx=-gR^2/x^2`

where *g *is the gravitational constant, *R *is the radius of the earth and *x *is the vertical distance travelled from the centre of the earth. Solve the differential equation to find the velocity of the particle in terms of height *x *subject to the initial condition that V= U when X= R.

Find the escape velocity of the particle (i.e. initial velocity to escape from the earth’s gravitational field) given that g = 9.81m/s2 and R = 6378 km. Express your answer in units of km/s.

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### 1 Answer

I will say that I hesitated in answering this question because V was not defined. Let's do this question as if `V = v(x)`. We get the following differential equation:

`v(dv)/(dx) = -(gR^2)/x^2`

This turns out to be a pretty good separable differential equation. Just multiply both sides by the differential `dx`:

`vdv = -gR^2*(dx)/x^2`

Now, we just integrate both sides:

`intvdv =-gR^2int(dx)/x^2`

The easiest way to solve this is to rearrange the integral on the right to reflect that we're just taking the integral of a power of x:

`intvdv = -gR^2intx^-2 dx`

Now, we integrate easily!

`v^2/2 + C_1 = -gR^2 (-x^-1 + C_2)`

Simplifying:

`v^2/2 = gR^2/x + C`

Keep in mind, we manipulated `C_2` by constants, including multiplying and by subtracting the other random constant of integration. We just put them all together into another random constant `C` for simplicity. Continuing, let's isolate `v` by multiplying by 2 and taking the square root:

`v = sqrt((2gR^2)/x + C)`

And there you have your function! We may have taken the negative square root if we know the velocity would be downward; however, we know that we throw it upward, so the velocity will be positive.

Now, we'll solve for the constant of integration C. We do this by using the initial condition:

`v(R) = U`

We simply substitute these values for x and v in the equation above and solve for C:

`U = sqrt((2gR^2)/R + C)`

We simplify by squaring both sides and subtracting 2gR (the result if we simplify the fraction by cancelling the R's):

`U^2 = 2gR + C`

`U^2 - 2gR = C`

And there is your constant of integration!

As a result, our final equation is as follows:

`v(x) = sqrt((2gR^2)/x + U^2 - 2gR)`

I hope that helps! Tell me if I mixed up what you meant by `V(dv)/(dx)`.

The escape velocity is calculated by finding U at the radius of the earth. Using Newton's Laws, we can calculate the escape velocity through the following equation (see link below):

`v_e = sqrt(2gR)`

Now, we would just put in the numbers you have given, but we don't have the right units! We need everything in terms of km and s. We need to change the units of g!

`g = 9.8 m/s^2 :- (1km)/(1000 m) = 0.0098 km/s^2`

Now, we can use the equation:

`v_e = sqrt(2(0.0098)(6738)) =11.5 (km)/s`

And there you have your answer!

**Sources:**