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How do you calculate the escape velocity of the sun with a given mass of 2 x 10 to the power of 30 kg and its radius is 7 x 10 to the power of 8 meters?
Wiki User
∙ 15y ago
A body in rest on the surface of sphere with given mass and radius has certain potential energy of Ep=-GMm / r. If you want to totally escape the body gravitational influence, you have to calculate how much kinetic energy you have to give the body for it to be able to reach a distance of infinity from the sphere's center, with at least 0 velocity left.
By the law of energy conservation, the initial energy must be equal to final energy:
Ep0 + Ek0 = Ep1 + Ek1
Ep0 = -GMm / r
Ek0 = mve2 / 2
Ep1 = -GMm / inf = 0
Ek1 = m * 0 / 2 = 0,
so:
-GMm / r + mve2 / 2 = 0,
simplyfying:
ve2 / 2 = GM / r,
then:
ve = sqrt(2GM/r)
Wiki User
∙ 15y ago
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How was the problem of escaping earth's gravity solved?
First you need to caluclate the escape volocity. Calculating an escape velocity 1. Determine the mass and radius of the planet you are on. For Earth, assuming that you are at sea level, the radius is 6.38x10^6 meters and the mass is 5.97x10^24 kilograms. You will need the gravitational constant (G), which is 6.67x10^-11 N m^2 kg^-2. It is recommended to use metric units. 2. Using the above data, calculate the required velocity needed to exceed the planet's gravitational force. The object must have greater energy than the planet's velocity as follows: V(escape)= squareroot[(2GM)/r] where "M" is the mass of the earth, "G" is the gravitational constant(6.67x10^-11) and "r" is the radius from the center of the planet(6.38x10^6). 3. Accelerate the mass to the escape velocity. It is optimal to accelerate it perpendicular to the ground, assuming it is level. Accelerating the mass at an angle other than 90 degrees with respect to the ground will require a greater velocity such that V(escape)=V(actual)*sin(theta), where theta is the angle between the ground and the projected acceleration vector. 4. The escape velocity of Earth comes to about 11.2 kilometers per second or 25000 miles per hour from the surface.
Is the critical velocity equal to orbital velocity of satellite?
Balance of force of gravity by centripetal forceYes, for any given orbital radius (r) only one velocity will give stable orbit, this is called the critical velocity.Pick your orbital radius and satellite mass (m)Use Gmm / r2 to calculate gravitational force (f),then find velocity from : v = sq root ( (r * f ) / m)
Why the moon does not escape from the earth?
They orbit around their common centre of gravity, the orbital radius and velocity of both (centripital force) is exactly enough to overcome the force of gravity between them.
The ratio of the radius of that of the moon is 10 .The ratio of g on the earth and on the moon is 6 then what is the ratio of the escape velocity from earth's surface to that from the moon's surface?
The escape velocity is given by √2gR Hence it's value Ve on the earth and Vm on the moon is Ve = √2ge.Re Vm = √2gm.Rm Therefore , their ratio = Ve/Vm = √ge.Re/√gm.Rm = √6 x 10 = √60 = 8 nearly
Radius of the sun in meters?
The radius of the Sun is approximates 695 500 kilometers.
The more massive is an object and the smaller is its radius the greater or smaller is the escape velocity from it?
For two bodies with equal radius, the more massive has the greater escape velocity. For two bodies with equal mass, the one with smaller radius has the greater escape velocity. Both conditions listed in the question indicate greaterescape velocity.
How do you convert radians per second to meters per second?
the tangential velocity is equal to the angular velocity multiplied by the radius the tangential velocity is equal to the angular velocity multiplied by the radius
What is the centripetal acceleration of an object being swung on a string with a radius of 3 meters at a velocity of 4 meters per second?
Use the formula for centripetal acceleration: velocity squared / radius.
What is the centripetal acceleration of an object being swung on a string with a radius of 5 meters at a velocity of 4 meters a second?
Use the formula a = v2 / r, with v = velocity (speed, actually) in meters/second, r = radius in meters. The answer will be in meters per square second.
What of a black hole is the radius from a black hole at which the escape velocity is approximately equal to the speed of light?
It is called the Schwarzschild radius
What occurs at the distance of the Schwarzschild radius from the center of a black hole?
The escape velocity reaches the speed of light at the Schwarzshild radius.
What is the velocity of a car if the wheel rotates 400 times a minute give your answer in meters per minute and kilometres per hour?
It is impossible to calculate this as we would need to know the radius of the car's wheel.
At what angular velocity would the earth have to rotate about its axis for a body at the equator to feel no weight?
You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.You can calculate it yourself, with the formula for acceleration in uniform circular motion. The centripetal acceleration, with circular motion, is v2 / r (velocity squared divided by the radius). Since the Earth's gravitation is about 9.8 meters per second square, solve the equation v2 / r = 9.8, for variable v. r (radius, of the Earth) should be converted to meters. The velocity will be in meters per second.Note that this exercise assumes the Earth is rigid. In practice, if Earth really rotated that fast, it would flatten out a lot, and in fact be torn apart.
How to calculate circumference of circle with radius of 2.250 meters?
Double the radius to 4.5 meters, which is the diameter. From there, multiply the diameter by Pi to get the circumference:4.5m x Pi = circumference of about 28.27 meters
Find the angular velocity of r equals 8.0?
Assuming that "r" is the radius, that simply isn't sufficient information to calculate angular velocity.
Is earth accelerating right now?
Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.
Derive a formula in term of acceleration due to gravity and radius for the Earth escape velocity?
m= mass of object v= velocity R= radius of the earth h= height from the surface to the escape velocity (mv2 ) ______ = G MEM R + h _____ (R+h)2 v2 =GME ______ (R +h)2 Ve = (√gR2 /(R+h)2 )
